10 INFINITE SERIES 10.1 Sequences Preliminary Questions 1. What is a4 for the sequence an = n2 − n? solution Substituting n = 4 in the expression for an gives a4 = 42 − 4 = 12. 2. Which of the following sequences converge to zero? (a) n2 n2 + 1 (b) 2n (c) −1 n 2 solution (a) This sequence does not converge to zero: lim n2 n→∞ n2 + 1 = lim x2 x→∞ x 2 + 1 = lim 1 x→∞ 1 + 1 x2 = 1 = 1. 1+0 (b) This sequence does not converge to zero: this is a geometric sequence with r = 2 > 1; hence, the sequence diverges to ∞. (c) Recall that if |an | converges to 0, then an must also converge to zero. Here, n n −1 = 1 , 2 2 which is a geometric sequence with 0 < r < 1; hence, ( 12 )n converges to zero. It therefore follows that (− 12 )n converges to zero. √ 3. Let an be the nth decimal approximation to 2. That is, a1 = 1, a2 = 1.4, a3 = 1.41, etc. What is lim an ? n→∞ solution √ lim an = 2. n→∞ 4. Which of the following sequences is defined recursively? √ (a) an = 4 + n (b) bn = 4 + bn−1 solution (a) an can be computed directly, since it depends on n only and not on preceding terms. Therefore an is defined explicitly and not recursively. (b) bn is computed in terms of the preceding term bn−1 , hence the sequence {bn } is defined recursively. 5. Theorem 5 says that every convergent sequence is bounded. Determine if the following statements are true or false and if false, give a counterexample. (a) If {an } is bounded, then it converges. (b) If {an } is not bounded, then it diverges. (c) If {an } diverges, then it is not bounded. solution (a) This statement is false. The sequence an = cos πn is bounded since −1 ≤ cos πn ≤ 1 for all n, but it does not converge: since an = cos nπ = (−1)n , the terms assume the two values 1 and −1 alternately, hence they do not approach one value. (b) By Theorem 5, a converging sequence must be bounded. Therefore, if a sequence is not bounded, it certainly does not converge. (c) The statement is false. The sequence an = (−1)n is bounded, but it does not approach one limit. 633 May 23, 2011 634 C H A P T E R 10 INFINITE SERIES Exercises 1. Match each sequence with its general term: a1 , a2 , a3 , a4 , . . . General term (a) 12 , 23 , 34 , 45 , . . . (i) cos πn (b) −1, 1, −1, 1, . . . n! (ii) n 2 (c) 1, −1, 1, −1, . . . (iii) (−1)n+1 (d) 21 , 24 , 68 , 24 16 . . . (iv) n n+1 solution (a) The numerator of each term is the same as the index of the term, and the denominator is one more than the numerator; n , n = 1, 2, 3, . . . . hence an = n+1 (b) The terms of this sequence are alternating between −1 and 1 so that the positive terms are in the even places. Since cos πn = 1 for even n and cos πn = −1 for odd n, we have an = cos πn, n = 1, 2, . . . . (c) The terms an are 1 for odd n and −1 for even n. Hence, an = (−1)n+1 , n = 1, 2, . . . (d) The numerator of each term is n!, and the denominator is 2n ; hence, an = 2n!n , n = 1, 2, 3, . . . . In Exercises 3–12, calculate the first four terms of the sequence, starting with n = 1. 1 Let ann = for n = 1, 2, 3, . . . . Write out the first three terms of the following sequences. 2n − 1 3 3. c(a) n =b = a (b) cn = an+3 nn! n+1 solution n = 1, 2, 3, 4 in the formula for cn gives (d) en = 2an − an+1 (c) dn =Setting an2 c1 = 3 31 = = 3, 1! 1 c3 = 27 9 33 = = , 3! 6 2 c2 = 9 32 = , 2! 2 c4 = 81 27 34 = = . 4! 24 8 5. a1 = 2, an+1 = 2an2 − 3 (2n − 1)! bn = For n = 1, 2, 3 we have: solution n! a2 = a1+1 = 2a12 − 3 = 2 · 4 − 3 = 5; a3 = a2+1 = 2a22 − 3 = 2 · 25 − 3 = 47; a4 = a3+1 = 2a32 − 3 = 2 · 2209 − 3 = 4415. The first four terms of {an } are 2, 5, 47, 4415. 7. bn = 5 + cos πn 1 1, nb= b1 = For n= solution 1, b2,n−1 3, 4+we have bn−1 b1 = 5 + cos π = 4; b2 = 5 + cos 2π = 6; b3 = 5 + cos 3π = 4; b4 = 5 + cos 4π = 6. The first four terms of {bn } are 4, 6, 4, 6. 1 1 1 9. cn = +2n+1 + ··· + cn 1=+(−1) 2 3 n solution c1 = 1; 3 1 = ; 2 2 3 1 11 1 1 ; =1+ + = + = 2 3 2 3 6 1 1 1 11 1 25 =1+ + + = + = . 2 3 4 6 4 12 c2 = 1 + c3 c4 May 23, 2011 S E C T I O N 10.1 Sequences 635 11. b1 = 2, b2 = 3, bn = 2bn−1 + bn−2 an = n + (n + 1) + (n + 2) + · · · + (2n) solution We need to find b3 and b4 . Setting n = 3 and n = 4 and using the given values for b1 and b2 we obtain: b3 = 2b3−1 + b3−2 = 2b2 + b1 = 2 · 3 + 2 = 8; b4 = 2b4−1 + b4−2 = 2b3 + b2 = 2 · 8 + 3 = 19. The first four terms of the sequence {bn } are 2, 3, 8, 19. 13. Find a formula for the nth term of each sequence. cn = n-place decimal approximation to e 1 −1 1 2 3 4 (a) , , , ... (b) , , , . . . 1 8 27 6 7 8 solution (a) The denominators are the third powers of the positive integers starting with n = 1. Also, the sign of the terms is alternating with the sign of the first term being positive. Thus, 1 (−1)1+1 a1 = 3 = ; 1 13 1 (−1)2+1 a2 = − 3 = ; 2 23 1 (−1)3+1 a3 = 3 = . 3 33 This rule leads to the following formula for the nth term: an = (−1)n+1 . n3 (b) Assuming a starting index of n = 1, we see that each numerator is one more than the index and the denominator is four more than the numerator. Thus, the general term an is an = n+1 . n+5 In Exercises 15–26, Theorem to determine the7.limit of the sequence or state that the sequence diverges. Suppose thatuse lim an = 41and lim bn = Determine: n→∞ n→∞ 12 (an + bn ) 15. a(a) n = lim (b) lim an3 n→∞ n→∞ solution We have an = f (n) where f (x) = 12; thus, (c) lim cos(πbn ) (d) lim (an2 − 2an bn ) n→∞ n→∞ lim an = lim f (x) = lim 12 = 12. n→∞ x→∞ x→∞ 5n − 1 17. bn = 4 = 20+−9 2 an 12n n 5x − 1 ; thus, solution We have bn = f (n) where f (x) = 12x + 9 lim 5n − 1 n→∞ 12n + 9 = lim 5x − 1 = x→∞ 12x + 9 5 . 12 19. cn = −2−n 4 + n − 3n2 an = We have solution c = f (n) where f (x) = −2−x ; thus, 4n2 + 1n 1 lim −2−n = lim −2−x = lim − x = 0. n→∞ x→∞ x→∞ 2 21. cn = 9n n 1 zn = We have cn = f (n) where f (x) = 9x ; thus, solution 3 lim 9n = lim 9x = ∞ n→∞ x→∞ Thus, the sequence 9n diverges. n 23. an = −1/n zn =n10 2+1 x ; thus, solution We have an = f (n) where f (x) = 2 x +1 lim n→∞ x 1 x 1 1 = lim = lim √ x = lim = lim = √ = 1. 2 2 x→∞ x→∞ x→∞ x→∞ x +1 1+0 x +1 n2 + 1 x2 + 1 1 + 12 2 n x May 23, 2011 x x 636 C H A P T E R 10 INFINITE SERIES 12nn + 2 25. an = an ln = −93+ 4n n +1 12x + 2 solution We have an = f (n) where f (x) = ln ; thus, −9 + 4x 12n + 2 12x + 2 12x + 2 = lim ln = ln lim = ln 3 lim ln n→∞ x→∞ x→∞ −9 + 4x −9 + 4n −9 + 4x In Exercises 27–30, use Theorem 4 to determine the limit of the sequence. rn = ln n − ln(n2 + 1) 1 27. an = 4 + n solution We have lim 4 + n→∞ Since 1 1 = lim 4 + = 4 n x→∞ x √ x is a continuous function for x > 0, Theorem 4 tells us that 1 1 √ lim 4 + = lim 4 + = 4 = 2 n→∞ n→∞ n n n3 = −1 e4n/(3n+9) an cos 29. an = 2n3 + 1 solution We have n3 lim n→∞ 2n3 + 1 = 1 2 Since cos−1 (x) is continuous for all x, Theorem 4 tells us that π n3 n3 −1 −1 lim lim cos = cos = cos−1 (1/2) = 3 3 n→∞ n→∞ 3 2n + 1 2n + 1 n 31. Let an = −1 . −n Find a number M such that: an = tan n + 1(e ) (a) |an − 1| ≤ 0.001 for n ≥ M. (b) |an − 1| ≤ 0.00001 for n ≥ M. Then use the limit definition to prove that lim an = 1. n→∞ solution (a) We have n − (n + 1) −1 n = 1 . − 1 = = |an − 1| = n+1 n + 1 n + 1 n + 1 1 ≤ 0.001, that is, n ≥ 999. It follows that we can take M = 999. Therefore |an − 1| ≤ 0.001 provided n+1 1 ≤ 0.00001, that is, n ≥ 99999. It follows that we can take M = 99999. (b) By part (a), |an − 1| ≤ 0.00001 provided n+1 We now prove formally that lim an = 1. Using part (a), we know that n→∞ |an − 1| = 1 < , n+1 provided n > 1 − 1. Thus, Let > 0 and take M = 1 − 1. Then, for n > M, we have |an − 1| = 1 1 < = . n+1 M +1 n 33. Use the limit definition to prove that lim n−2 = 0. n→∞ Let bn = 13 . solution see that (a) FindWe a value of M such that |bn | ≤ 10−5 for n ≥ M. (b) Use the limit definition to prove that lim bn = 0. 1 1 −2 − 0| = n→∞ |n n2 = n2 < May 23, 2011 S E C T I O N 10.1 Sequences 637 provided 1 n> √ . Thus, let > 0 and take M = √1 . Then, for n > M, we have 1 1 1 |n−2 − 0| = 2 = 2 < 2 = . n n M In Exercises 35–62, use the appropriate limit laws andn theorems to determine the limit of the sequence or show that it = 1. Use the limit definition to prove that lim diverges. n→∞ n + n−1 1 n 35. an = 10 + − 9 solution By the Limit Laws for Sequences we have: 1 n 1 n 1 n lim 10 + − = 10 + lim − . = lim 10 + lim − n→∞ n→∞ n→∞ n→∞ 9 9 9 Now, n n 1 n 1 1 ≤ − ≤ . 9 9 9 − Because n 1 = 0, n→∞ 9 lim by the Limit Laws for Sequences, n n 1 1 = − lim = 0. n→∞ n→∞ 9 9 lim − Thus, we have 1 n = 0, lim − n→∞ 9 and 1 n 10 + − = 10 + 0 = 10. n→∞ 9 lim 37. cn = 1.01√n √ dn = n + 3 − n solution Since cn = f (n) where f (x) = 1.01x , we have lim 1.01n = lim 1.01x = ∞ n→∞ x→∞ so that the sequence diverges. 39. an = 21/n 2 bn = e1−n solution Because 2x is a continuous function, lim 21/n = lim 21/x = 2limx→∞ (1/x) = 20 = 1. n→∞ x→∞ 9n 41. cn = = n1/n bn n! solution For n ≥ 9, write cn = 9 9 9 9 9 9 9 9n = · ··· · · ··· · n! 1 2 9 10 11 n − 1 n call this C Each factor is less than 1 Then clearly 0≤ May 23, 2011 9n 9 ≤C n! n 638 C H A P T E R 10 INFINITE SERIES since each factor after the first nine is < 1. The squeeze theorem tells us that 9 9n 9 ≤ lim C = C lim =C·0=0 n→∞ n! n→∞ n n→∞ n lim 0 ≤ lim n→∞ so that limn→∞ cn = 0 as well. 3n2 + n+2 43. an = 82n an =2n2 − 3 n! solution 3n2 + n + 2 3x 2 + x + 2 3 = lim = . 2 n→∞ 2n − 3 x→∞ 2x 2 − 3 2 lim cos n √ 45. an = n an =n √ n + 4 ≤ cos n ≤ 1 the following holds: solution Since −1 − cos n 1 1 ≤ ≤ . n n n We now apply the Squeeze Theorem for Sequences and the limits lim − n→∞ 1 1 = lim =0 n n→∞ n to conclude that lim cosn n = 0. n→∞ n 47. dn = ln 5 − ln n! (−1)n cn = Note √ that solution n dn = ln 5n n! so that edn = 5n n! so lim edn = lim n→∞ 5n n→∞ n! =0 by the method of Exercise 41. If dn converged, we could, since f (x) = ex is continuous, then write lim edn = elimn→∞ dn = 0 n→∞ which is impossible. Thus {dn } diverges. 4 1/3 dn =2 ln(n + 22 + 4) − ln(n2 − 1) 49. an = n 1/3 solution Let an = 2 + 42 . Taking the natural logarithm of both sides of this expression yields n 4 4 1/3 1 = ln 2 + 2 . ln an = ln 2 + 2 3 n n Thus, 4 1/3 1 4 1 4 1 ln 2 + 2 lim ln 2 + 2 = ln lim 2 + 2 = n→∞ 3 x→∞ 3 x→∞ 3 n x x lim ln an = lim n→∞ = 1 1 ln (2 + 0) = ln 2 = ln 21/3 . 3 3 Because f (x) = ex is a continuous function, it follows that 1/3 lim an = lim eln an = elimn→∞ (ln an ) = eln 2 n→∞ n→∞ = 21/3 . 2n + 1 51. cn = ln 2 3n−1 + 41 − bn = tan n solution Because f (x) = ln x is a continuous function, it follows that 2 2x + 1 2x + 1 = ln lim = ln . lim cn = lim ln n→∞ x→∞ x→∞ 3x + 4 3x + 4 3 May 23, 2011 S E C T I O N 10.1 Sequences 639 en 53. yn = cn 2=n n nn + n1/nn n solution 2e n = 2e and 2e > 1. By the Limit of Geometric Sequences,we conclude that limn→∞ 2e = ∞. Thus, the given sequence diverges. en +n (−3)n 55. yn = an = n5n 2 solution e n −3 n en + (−3)n = lim + lim lim n→∞ n→∞ 5 n→∞ 5n 5 assuming both limits on the right-hand side exist. But by the Limit of Geometric Sequences, since −1 < −3 e <0< <1 5 5 both limits on the right-hand side are 0, so that yn converges to 0. π 57. an = n sin n 3 −n (−1) n n +2 bn = + 4−n on Sequences Defined by a Function, we have solution By 3n the3 Theorem lim n sin n→∞ Now, π π = lim x sin . x→∞ n x cos πx − π2 sin πx π π x lim x sin = lim = lim = lim π cos x→∞ x→∞ 1 x→∞ x→∞ x x − 12 x x π = π lim cos = π cos 0 = π · 1 = π. x→∞ x Thus, lim n sin n→∞ π = π. n 3 − 4n n! 59. bn = bn 2=+ 7n· 4n π solution Divide the numerator and denominator by 4n to obtain n 4 3 3 3 − 4n 4n − 4n = 4n − 1 . an = = n 2 7·4 2 2 + 7 · 4n 4n + 4n 4n + 7 Thus, 3 −1 limx→∞ 43x − 1 3 limx→∞ x 4 = lim an = lim 2 = n→∞ x→∞ 2 2 limx→∞ limx→∞ 4x + 7 4x + 7 1 4x − limx→∞ 1 = 3 · 0 − 1 = − 1 . 1 2·0+7 7 4x − limx→∞ 7 1 n 61. an = 1 +3 − 4n an = n 2 + 7 · 3n solution Taking the natural logarithm of both sides of this expression yields n ln 1 + 1 n 1 1 ln an = ln 1 + = = n ln 1 + . 1 n n n Thus, lim (ln an ) = lim n→∞ ln 1 + x1 x→∞ 1 x = lim x→∞ d dx 1 · − 1 1 2 ln 1 + x1 x 1+ x 1 1 = lim = 1. = lim = 1 1 x→∞ x→∞ d 1 1 + 0 1+ x − 2 dx x x Because f (x) = ex is a continuous function, it follows that lim an = lim eln an = elimn→∞ (ln an ) = e1 = e. n→∞ 1 n an = 1 + 2 n May 23, 2011 n→∞ 640 C H A P T E R 10 INFINITE SERIES In Exercises 63–66, find the limit of the sequence using L’Hôpital’s Rule. (ln n)2 n solution 63. an = d (ln x)2 2 ln x (ln n)2 (ln x)2 2 ln x x = lim = lim = lim dx d = lim n→∞ x→∞ x→∞ x→∞ x→∞ n x 1 x dx x lim d 2 ln x 2 2 = lim x = lim = lim dx d =0 x→∞ x→∞ x→∞ 1 x dx x 65. cn = n √n2 + 1 − n1 bn = n ln 1 + solution n x x2 + 1 − x x2 + 1 + x lim n n2 + 1 − n = lim x x 2 + 1 − x = lim n→∞ x→∞ x→∞ x2 + 1 + x = lim x→∞ = lim x→∞ x x2 + 1 + x 1+ 1 x2 x 2 +1 d x 1 = lim dx x→∞ d x 2 + 1 + x x→∞ 1 + √ x dx x 2 +1 = lim = lim x→∞ 1+ 1 1 1+(1/x 2 ) = 1 2 In Exercises 67–70, Theorem to evaluate lim an by verifying the given inequality. 3 use the Squeeze n→∞ dn = n2 n3 + 1 − n 1 1 1 67. an = , √ ≤ an ≤ √ 4 2 4 8 2n 2n n +n 4 8 solution For all n > 1 we have n < n , so the quotient √ 41 8 is smaller than √ 41 4 and larger than √ 81 n +n That is, n +n n +n8 . 1 1 1 an < = √ ; and = √ 2n2 n4 · 2 n4 + n4 1 1 1 = √ = √ . an > 4 8 8 8 2n 2n n +n Now, since lim √ n→∞ 1 2n4 1 = lim √ = 0, the Squeeze Theorem for Sequences implies that lim an = 0. n→∞ 2n2 n→∞ 1/n · 3 69. an = (2n + 31n )1/n , 3 ≤1an ≤ (2 · 3n )1/n = 1 2 n n n c = + + · · · + , n solution Clearly 2 + 3 2≥ 3 for all n ≥ 1.2Therefore: n2 + 1 n +2 n +n n n (2n + 3n )1/n ≥ (3n )1/n = 3. ≤ cn ≤ 2 2 n +n n +1 Also 2n + 3n ≤ 3n + 3n = 2 · 3n , so (2n + 3n )1/n ≤ (2 · 3n )1/n = 21/n · 3. Thus, 3 ≤ (2n + 3n )1/n ≤ 21/n · 3. Because lim 21/n · 3 = 3 lim 21/n = 3 · 1 = 3 n→∞ n→∞ and limn→∞ 3 = 3, the Squeeze Theorem for Sequences guarantees lim (2n + 3n )1/n = 3. n→∞ ofn the isn equivalent to the assertion lim an = L? Explain. 1/n (n + 10 )1/nfollowing , 10 ≤ astatements an =Which n ≤ (2 · 10 ) n→∞ (a) For every > 0, the interval (L − , L + ) contains at least one element of the sequence {an }. (b) For every > 0, the interval (L − , L + ) contains all but at most finitely many elements of the sequence {an }. solution Statement (b) is equivalent to Definition 1 of the limit, since the assertion “|an − L| < for all n > M” means that L − < an < L + for all n > M; that is, the interval (L − , L + ) contains all the elements an except (maybe) the finite number of elements a1 , a2 , . . . , aM . 71. May 23, 2011 S E C T I O N 10.1 Sequences 641 Statement (a) is not equivalent to the assertion lim an = L. We show this, by considering the following sequence: n→∞ ⎧ 1 ⎪ ⎪ ⎨ n an = ⎪ ⎪ ⎩1 + 1 n for odd n for even n Clearly for every > 0, the interval (−, ) = (L − , L + ) for L = 0 contains at least one element of {an }, but the sequence diverges (rather than converges to L = 0). Since the terms in the odd places converge to 0 and the terms in the even places converge to 1. Hence, an does not approach one limit. 3n2 1 73. Show thatthat an = Find an upper bound. is decreasing. Show an n=2 + 2 is increasing. 2n + 1 solution 2 . Then Let f (x) = 3x 2 x +2 f (x) = 6x(x 2 + 2) − 3x 2 · 2x 2 (x 2 + 2) = 12x 2 (x 2 + 2) . f (x) > 0 for x > 0, hence f is increasing on this interval. It follows that an = f (n) is also increasing. We now show that M = 3 is an upper bound for an , by writing: 3n2 3n2 + 6 3(n2 + 2) ≤ 2 = = 3. an = 2 n +2 n +2 n2 + 2 That is, an ≤ 3 for all n. 75. Give an example of √ a3 divergent sequence {an } such that lim |an | converges. n→∞ Show that an = n + 1 − n is decreasing. solution Let an = (−1)n . The sequence {an } diverges because the terms alternate between +1 and −1; however, the sequence {|an |} converges because it is a constant sequence, all of whose terms are equal to 1. 77. Using the limit definition, prove that if {an } converges and {bn } diverges, then {an + bn } diverges. Give an example of divergent sequences {an } and {bn } such that {an + bn } converges. solution We will prove this result by contradiction. Suppose limn→∞ an = L1 and that {an + bn } converges to a limit L2 . Now, let > 0. Because {an } converges to L1 and {an + bn } converges to L2 , it follows that there exist numbers M1 and M2 such that: 2 | (an + bn ) − L2 | < 2 |an − L1 | < for all n > M1 , for all n > M2 . Thus, for n > M = max{M1 , M2 }, |an − L1 | < 2 and | (an + bn ) − L2 | < . 2 By the triangle inequality, |bn − (L2 − L1 )| = |an + bn − an − (L2 − L1 )| = |(−an + L1 ) + (an + bn − L2 )| ≤ |L1 − an | + |an + bn − L2 |. Thus, for n > M, |bn − (L2 − L1 ) | < + = ; 2 2 that is, {bn } converges to L2 − L1 , in contradiction to the given data. Thus, {an + bn } must diverge. 79. Theorem 1 states that if lim f (x) = L, then the sequence an = f (n) converges and lim an = L. Show that the x→∞ n→∞ limit L, then there exists a Use the limit definition to prove that if {an } is a convergent sequence of integers with converse is false. In other find a function f (x) such that an = f (n) converges but lim f (x) does not exist. number M such that awords, n = L for all n ≥ M. x→∞ solution Let f (x) = sin πx and an = sin πn. Then an = f (n). Since sin πx is oscillating between −1 and 1 the limit lim f (x) does not exist. However, the sequence {an } is the constant sequence in which an = sin π n = 0 for all n, x→∞ hence it converges to zero. Use the limit definition to prove that the limit does not change if a finite number of terms are added or removed from a convergent sequence. May 23, 2011 642 C H A P T E R 10 INFINITE SERIES 81. Let bn = an+1 . Use the limit definition to prove that if {an } converges, then {bn } also converges and lim an = n→∞ lim bn . n→∞ solution Suppose {an } converges to L. Let bn = an+1 , and let > 0. Because {an } converges to L, there exists an M such that |an − L| < for n > M . Now, let M = M − 1. Then, whenever n > M, n + 1 > M + 1 = M . Thus, for n > M, |bn − L| = |an+1 − L| < . Hence, {bn } converges to L. √ that lim a exists if and only if there nonzero. Show Let {an } be a sequence such that lim |an | exists√and is √ n increasing and bounded above by 83. Proceed as in Example 12 to show that the sequence 3, 3 3, 3 3 3, . . . isn→∞ n→∞ does not change for n > M. exists an integer M such that the sign of a n its value. M = 3. Then prove that the limit exists and find solution This sequence is defined recursively by the formula: an+1 = 3an , a1 = √ 3. Consider the following inequalities: √ √ √ 3a1 = 3 3 > 3 = a1 √ √ a3 = 3a2 > 3a1 = a2 √ √ a4 = 3a3 > 3a2 = a3 a2 = ⇒ a2 > a1 ; ⇒ a3 > a2 ; ⇒ a4 > a3 . In general, if we assume that ak > ak−1 , then ak+1 = 3ak > 3ak−1 = ak . Hence, by mathematical {an } is increasing. √ √ induction, an+1 > an for all n; that is, the sequence Because an+1 = 3an , it follows that an ≥ 0 for all n. Now, a1 = 3 < 3. If ak ≤ 3, then √ ak+1 = 3ak ≤ 3 · 3 = 3. Thus, by mathematical induction, an ≤ 3 for all n. Since {an } is increasing and bounded, it follows by the Theorem on Bounded Monotonic Sequences that this sequence is converging. Denote the limit by L = limn→∞ an . Using Exercise 81, it follows that √ L = lim an+1 = lim 3an = 3 lim an = 3L. n→∞ n→∞ n→∞ √ Thus, L2 = 3L, so L =√0 or L = 3. Because the sequence is increasing, we have an ≥ a1 = 3 for all n. Hence, the limit also satisfies L ≥ 3. We conclude that the appropriate solution is L = 3; that is, lim an = 3. n→∞ Let {an } be the sequence defined recursively by Further Insights and Challenges √ n n/2 by observing that half of the factors of n! are greater n! = ∞. Hint: Verify that n!0,≥ (n/2) a0 = an+1 = 2 + an n→∞ than or equal to n/2. √ √ n √ n/2 Thus, a1 We = show 2, that a2 =n! ≥2 + 2,. For a3 n=≥ 42even, + 2we + have: 2, . . . . solution 2 (a) Show that if an < 2, then an+1 < 2. Conclude by induction that 2 for all n. n an < n n + 1 by· ·induction · · · n ≥ that+{a1n } ·is· ·increasing. · · n. · n! a=n 1≤· a· n+1 · · · . Conclude (b) Show that if an < 2, then √ L= 2lim 2a exists. 2 Then compute by showing (c) Use (a) and (b) to conclude that L that L = 2 + L. nn n n 85. Show that lim 2 factorsn→∞ 2 factors 2 factors Since each one of the n2 factors is greater than n2 , we have: n! ≥ n n n n/2 n + 1 · ··· · n ≥ · ··· · = . 2 2 2 2 n 2 n 2 factors factors For n ≥ 3 odd, we have: n+1 n−1 n+1 · ··· · n ≥ · · · · · n. · n! = 1 · · · · · 2 2 2 n−1 2 May 23, 2011 factors n+1 2 factors n+1 2 factors S E C T I O N 10.1 Sequences 643 n Since each one of the n+1 2 factors is greater than 2 , we have: n! ≥ n n (n+1)/2 n n/2 n n n/2 n n+1 · ··· · n ≥ · ··· · = ≥ = . 2 2 2 2 2 2 2 n+1 2 factors n+1 2 factors n/2 . Thus, In either case we have n! ≥ n2 √ n n! ≥ Since lim n→∞ n = ∞, it follows that lim 2 n→∞ n . 2 √ √ n n! = ∞. Thus, the sequence an = n n! diverges. 87. Given positive√ numbers a1 < b1 , define two sequences recursively by n n! . Let bn = an + bn n an+1 = an bn , bn+1 = n 2 1 k = ln . (a) Show that ln b n (a) Show that an ≤ bn for all n n (Figure n 13). k=1and {b } is decreasing. (b) Show that {an } is increasing n bn − an 1 converges to (b) Show that ln b (c) Show that bn+1 − nan+1 ≤ . ln x dx, and conclude that bn → e−1 . 2 0 (d) Prove that both {an } and {bn } converge and have the same limit. This limit, denoted AGM(a1 , b1 ), is called the arithmetic-geometric √ mean of a1 and b1 . (e) Estimate AGM(1, 2) to three decimal places. Geometric Arithmetic mean mean an a n+1 b n+1 AGM(a 1, b 1) bn x FIGURE 13 solution (a) Examine the following: √ 2 √ 2 √ √ √ an − 2 an bn + bn an + bn an + bn − 2 an bn bn+1 − an+1 = − an bn = = 2 2 2 √ √ 2 an − bn ≥ 0. = 2 We conclude that bn+1 ≥ an+1 for all n > 1. By the given information b1 > a1 ; hence, bn ≥ an for all n. (b) By part (a), bn ≥ an for all n, so √ an+1 = an bn ≥ an · an = an2 = an for all n. Hence, the sequence {an } is increasing. Moreover, since an ≤ bn for all n, bn+1 = bn + bn 2bn an + bn ≤ = = bn 2 2 2 for all n; that is, the sequence {bn } is decreasing. (c) Since {an } is increasing, an+1 ≥ an . Thus, bn+1 − an+1 ≤ bn+1 − an = an + bn an + bn − 2an bn − an − an = = . 2 2 2 Now, by part (a), an ≤ bn for all n. By part (b), {bn } is decreasing. Hence bn ≤ b1 for all n. Combining the two inequalities we conclude that an ≤ b1 for all n. That is, the sequence {an } is increasing and bounded (0 ≤ an ≤ b1 ). By the Theorem on Bounded Monotonic Sequences we conclude that {an } converges. Similarly, since {an } is increasing, an ≥ a1 for all n. We combine this inequality with bn ≥ an to conclude that bn ≥ a1 for all n. Thus, {bn } is decreasing and bounded (a1 ≤ bn ≤ b1 ); hence this sequence converges. To show that {an } and {bn } converge to the same limit, note that − an−1 −a b b b −a bn − an ≤ n−1 ≤ n−2 2 n−2 ≤ · · · ≤ 1 n−1 1 . 2 2 2 Thus, lim (bn − an ) = (b1 − a1 ) lim n→∞ May 23, 2011 1 n→∞ 2n−1 = 0. 644 C H A P T E R 10 INFINITE SERIES (d) We have an+1 = an bn , a1 = 1; bn+1 = an + bn , 2 b1 = √ 2 Computing the values of an and bn until the first three decimal digits are equal in successive terms, we obtain: √ a2 = a1 b1 = 1 · 2 = 1.1892 √ 1+ 2 a1 + b1 b2 = = = 1.2071 2 2 √ a3 = a2 b2 = 1.1892 · 1.2071 = 1.1981 a + b2 1.1892 · 1.2071 b3 = 2 = = 1.1981 2 2 a4 = a3 b3 = 1.1981 a + b3 b4 = 3 = 1.1981 2 Thus, √ AGM 1, 2 ≈ 1.198. Let a1n = Hn1− ln n, where Hn is the 1 1 nth harmonic number + + ··· + . Let cn = + n n+1 n+2 2n 1 1 1 (a) Calculate c1 , c2 , c3 , c4 . Hn = 1 + + + · · · + 2 y3= x −1 over n the interval [n, 2n] to prove that (b) Use a comparison of rectangles with the area under n+1 2n dx 2n dx dx 1 (a) Show that an ≥ 0 for n ≥ 1. Hint: Show that . Hn ≥ 1 + 1 ≤ cn x≤ + x 2n x n n n (b) Show that {an } is decreasing by interpreting an − an+1 as an area. (c) Use thelim Squeeze Theorem to determine lim cn . an exists. (c) Prove that n→∞ 89. n→∞ This limit, denoted γ , is known as Euler’s Constant. It appears in many areas of mathematics, including analysis and number theory, and has been calculated to more than 100 million decimal places, but it is still not known whether γ is an irrational number. The first 10 digits are γ ≈ 0.5772156649. solution (a) Since the function y = x1 is decreasing, the left endpoint approximation to the integral 1n+1 dx x is greater than this integral; that is, 1·1+ n+1 1 1 1 dx · 1 + · 1 + ··· + · 1 ≥ 2 3 n x 1 or Hn ≥ n+1 dx . x 1 y 1 1 1 2 3 1 2 3 1/n n n+1 Moreover, since the function y = x1 is positive for x > 0, we have: n+1 n dx dx ≥ . x 1 1 x May 23, 2011 x S E C T I O N 10.2 Summing an Infinite Series 645 Thus, Hn ≥ n n dx = ln x = ln n − ln 1 = ln n, 1 1 x and an = Hn − ln n ≥ 0 for all n ≥ 1. (b) To show that {an } is decreasing, we consider the difference an − an+1 : an − an+1 = Hn − ln n − Hn+1 − ln(n + 1) = Hn − Hn+1 + ln(n + 1) − ln n 1 1 1 1 1 + ln(n + 1) − ln n = 1 + + ··· + − 1 + + ··· + + 2 n 2 n n+1 =− 1 + ln(n + 1) − ln n. n+1 n+1 dx 1 Now, ln(n + 1) − ln n = nn+1 dx x , whereas n+1 is the right endpoint approximation to the integral n x . Recalling 1 y = x is decreasing, it follows that n+1 1 dx ≥ x n + 1 n y y= 1 x 1 n+1 n+1 n x so an − an+1 ≥ 0. (c) By parts (a) and (b), {an } is decreasing and 0 is a lower bound for this sequence. Hence 0 ≤ an ≤ a1 for all n. A monotonic and bounded sequence is convergent, so limn→∞ an exists. 10.2 Summing an Infinite Series Preliminary Questions 1. What role do partial sums play in defining the sum of an infinite series? solution The sum of an infinite series is defined as the limit of the sequence of partial sums. If the limit of this sequence does not exist, the series is said to diverge. 2. What is the sum of the following infinite series? 1 1 1 1 1 + + + + + ··· 4 8 16 32 64 solution This is a geometric series with c = 14 and r = 12 . The sum of the series is therefore 1 4 1 1 = 41 = . 1 2 1− 2 2 3. What happens if you apply the formula for the sum of a geometric series to the following series? Is the formula valid? 1 + 3 + 32 + 33 + 34 + · · · May 23, 2011 646 C H A P T E R 10 INFINITE SERIES solution This is a geometric series with c = 1 and r = 3. Applying the formula for the sum of a geometric series then gives ∞ 3n = n=0 1 1 =− . 1−3 2 Clearly, this is not valid: a series with all positive terms cannot have a negative sum. The formula is not valid in this case because a geometric series with r = 3 diverges. ∞ 1 1 4. Arvind asserts that = 0 because 2 tends to zero. Is this valid reasoning? n2 n n=1 solution Arvind’s reasoning is not valid. Though the terms in the series do tend to zero, the general term in the sequence of partial sums, 1 1 1 Sn = 1 + 2 + 2 + · · · + 2 , 2 3 n is clearly larger than 1. The sum of the series therefore cannot be zero. ∞ 1 5. Colleen claims that √ converges because n n=1 1 lim √ = 0 n→∞ n Is this valid reasoning? solution Colleen’s reasoning is not valid. Although the general term of a convergent series must tend to zero, a series ∞ 1 whose general term tends to zero need not converge. In the case of √ , the series diverges even though its general n n=1 term tends to zero. 6. Find an N such that SN > 25 for the series ∞ 2. n=1 solution The Nth partial sum of the series is: SN = N n=1 2 = 2 + · · · + 2 = 2N. N ∞ 7. Does there exist an N such that SN > 25 for the series solution The series ∞ 2−n ? Explain. n=1 2−n is a convergent geometric series with the common ratio r = n=1 S= 1 . The sum of the series is: 2 1 2 = 1. 1 − 12 Notice that the sequence of partial sums {SN } is increasing and converges to 1; therefore SN ≤ 1 for all N. Thus, there does not exist an N such that SN > 25. 8. Give an example of a divergent infinite series whose general term tends to zero. solution Consider the series ∞ 1 9 n=1 n 10 . The general term tends to zero, since lim n→∞ 1 9 n 10 = 0. However, the Nth partial sum satisfies the following inequality: SN = 1 9 1 10 + 1 1 9 2 10 1 + ··· + 1 N 9 10 ≥ N N 9 10 9 1 = N 1− 10 = N 10 . That is, SN ≥ N 10 for all N. Since lim N 10 = ∞, the sequence of partial sums Sn diverges; hence, the series N→∞ diverges. May 23, 2011 ∞ 1 9 n=1 n 10 S E C T I O N 10.2 Summing an Infinite Series Exercises 1. Find a formula for the general term an (not the partial sum) of the infinite series. 1 1 1 1 1 5 25 125 (a) + + + + ··· (b) + + + + ··· 3 9 27 81 1 2 4 8 1 22 33 44 − + − + ··· 1 2·1 3·2·1 4·3·2·1 2 1 2 1 (d) 2 + 2 + 2 + 2 + ··· 1 +1 2 +1 3 +1 4 +1 (c) solution (a) The denominators of the terms are powers of 3, starting with the first power. Hence, the general term is: 1 an = n . 3 (b) The numerators are powers of 5, and the denominators are the same powers of 2. The first term is a1 = 1 so, an = n−1 5 . 2 (c) The general term of this series is, an = (−1)n+1 nn . n! (d) Notice that the numerators of an equal 2 for odd values of n and 1 for even values of n. Thus, ⎧ 2 ⎪ ⎪ ⎨ n2 + 1 odd n an = ⎪ ⎪ ⎩ 1 even n n2 + 1 The formula can also be rewritten as follows: n+1 1 + (−1) 2 +1 . an = n2 + 1 In Exercises compute the partial sums S2 , S4 , and S6 . Write3–6, in summation notation: 1 1 1 1 +1 + · · · (a) 11+ + 3. 1 + 2 +4 2 9+ 216+ · · · 2 3 4 1 1 1 (c) 1 − + − + · · · solution 3 5 7 1 5 125 625 3125 15,625 S2 =+1·+ + + + ·· 2 = 4; (d) 2 9 16 25 36 (b) 1 1 1 1 + + + + ··· 9 16 25 36 1 1 1 205 ; S4 = 1 + 2 + 2 + 2 = 144 2 3 4 1 1 1 1 1 5369 . S6 = 1 + 2 + 2 + 2 + 2 + 2 = 3600 2 3 4 5 6 1 1 1 ∞+ + + ··· 1 · 2 (−1) 2 · 3k k −13 · 4 solution k=1 5. S2 = 1 1 1 4 2 1 + = + = = ; 1·2 2·3 2 6 6 3 S4 = S2 + a3 + a4 = 2 1 1 2 1 1 4 + + = + + = ; 3 3·4 4·5 3 12 20 5 S6 = S4 + a5 + a6 = 1 1 4 1 1 6 4 + + = + + = . 5 5·6 6·7 5 30 42 7 ∞ 1 j! j =1 May 23, 2011 647 648 C H A P T E R 10 INFINITE SERIES 2 3 7. The series S = 1 + 15 + 15 + 15 + · · · converges to 54 . Calculate SN for N = 1, 2, . . . until you find an SN that approximates 54 with an error less than 0.0001. solution S1 = 1 S2 = 1 + S3 = 1 + S3 = 1 + S4 = 1 + S5 = 1 + 1 5 1 5 1 5 1 5 1 5 = + + + + 6 = 1.2 5 1 31 = = 1.24 25 25 1 1 156 + = = 1.248 25 125 125 1 1 1 781 + + = = 1.2496 25 125 625 625 1 1 1 1 3906 + + + = = 1.24992 25 125 625 3125 3125 Note that 1.25 − S5 = 1.25 − 1.24992 = 0.00008 < 0.0001 In Exercises 9 and 10, use system to compute S10 , S100 , S500 , and S1000 for the series. Do these 1 1 algebra 1 1 a computer −to the +given − value? + · · · is known to converge to e−1 (recall that 0! = 1). Calculate SN for The series S= values suggest convergence 0! 1! 2! 3! N = 1, 2, . . . until you find an SN that approximates e−1 with an error less than 0.001. 9. 1 1 1 1 π −3 = − + − + ··· 4 2 · 3 · 4 4 · 5 · 6 6 · 7 · 8 8 · 9 · 10 solution Write an = (−1)n+1 2n · (2n + 1) · (2n + 2) Then SN = N an i=1 Computing, we find π −3 ≈ 0.0353981635 4 S10 ≈ 0.03535167962 S100 ≈ 0.03539810274 S500 ≈ 0.03539816290 S1000 ≈ 0.03539816334 It appears that SN → π−3 4 . 11. Calculate S3 , S4 , and S5 and then find the sum of the telescoping series ∞ 1 1 − Sπ 4= 1 1 1 1 n++ 2 + · · · = 1 + n ++ 90 n=1 24 34 44 solution 1 1 1 1 1 1 3 1 1 − + − + − = − = ; 2 3 3 4 4 5 2 5 10 1 1 1 1 1 − = − = ; S4 = S3 + 5 6 2 6 3 1 1 5 1 1 − = − = . S5 = S4 + 6 7 2 7 14 S3 = May 23, 2011 S E C T I O N 10.2 Summing an Infinite Series The general term in the sequence of partial sums is 1 1 1 1 1 1 1 1 1 1 SN = − + − + − + ··· + − = − ; 2 3 3 4 4 5 N +1 N +2 2 N +2 thus, 1 1 − N +2 N→∞ 2 S = lim SN = lim N→∞ = 1 . 2 The sum of the telescoping series is therefore 12 . ∞ 1 13. Calculate S3 , S4 , and S5 and then find the sum S = using the identity 2−1 ∞ 4n 1 n=1 as a telescoping series and find its sum. Write 1 1 1 1 n(n − 1) − = n=3 2 2n − 1 2n + 1 4n2 − 1 solution 1 1 1 1 1 1 1 1 1 1 3 − + − + − = 1− = ; 1 3 2 3 5 2 5 7 2 7 7 1 1 4 1 1 1 − = 1− = ; S4 = S3 + 2 7 9 2 9 9 1 1 1 5 1 1 − = 1− = . S5 = S4 + 2 9 11 2 11 11 1 S3 = 2 The general term in the sequence of partial sums is 1 1 1 1 1 1 1 1 1 1 1 1 1 1 SN = − + − + − + ··· + − = 1− ; 2 1 3 2 3 5 2 5 7 2 2N − 1 2N + 1 2 2N + 1 thus, 1 1 1 1− = . 2N + 1 2 N→∞ 2 S = lim SN = lim N→∞ 1 1 1 ∞+ ··· . + + 15. Find the sum of 1 1 · 3 3 · 5 5 Use partial fractions to rewrite · 7 as a telescoping series and find its sum. n(n + 3) solution We may write this sum asn=1 ∞ n=1 ∞ 1 1 1 1 = − . (2n − 1)(2n + 1) 2 2n − 1 2n + 1 n=1 The general term in the sequence of partial sums is 1 1 1 1 1 1 1 1 1 1 1 1 1 1 SN = − + − + − + ··· + − = 1− ; 2 1 3 2 3 5 2 5 7 2 2N − 1 2N + 1 2 2N + 1 thus, 1 1 1 1− = , 2N + 1 2 N→∞ 2 lim SN = lim N→∞ and ∞ n=1 1 1 = . (2n − 1)(2n + 1) 2 In Exercises 17–22, use Theorem 3 to prove that the ∞ following series diverge. (−1)n−1 and show that the series diverges. Find a formula for the partial sum SN of ∞ n n=1 17. 10n + 12 n=1 solution The general term, n , has limit 10n + 12 lim n n→∞ 10n + 12 = lim Since the general term does not tend to zero, the series diverges. May 23, 2011 1 n→∞ 10 + (12/n) = 1 10 649 650 C H A P T E R 10 INFINITE SERIES 0 1 2 3 ∞ + − + ··· − 1 2 3n 4 n−1 solution general term an = (−1)n−1 n−1 n2 + 1 n does not tend to zero. In fact, because limn→∞ n = 1, limn→∞ an n=1 The does not exist. By Theorem 3, we conclude that the given series diverges. 1 1 1 ∞+ cos + cos + · · · 21. cos 2 (−1)n3n2 4 1 solution n=1 The general term an = cos n+1 tends to 1, not zero. By Theorem 3, we conclude that the given series diverges. 19. In Exercises 23–36, use the formula for the sum of a geometric series to find the sum or state that the series diverges. ∞ 2 1 +1−n 1 1 4n 23. +n=0 + 2 + · · · 1 8 8 solution This is a geometric series with c = 1 and r = 18 , so its sum is 1 8 1 = 7/8 7 = 1 − 18 ∞ −n 3 4 4 45 43 + + + ··· 11 n=353 55 54 solution Rewrite this series as 25. ∞ n 11 3 n=3 11 > 1, so it is divergent. This is a geometric series with r = 3 ∞ 4 n ∞ − 27. 7 ·9(−3)n n=−4 5n n=2 4 solution This is a geometric series with c = 1 and r = − , starting at n = −4. Its sum is thus 9 c 1 95 59,049 cr −4 = 4 = 4 = = 5 5 4 + 45 4 + 4 1−r 3328 r −r 9 · 4 5 4 9 9 ∞ ∞−n 29. e π n n=1 e n=0 solution Rewrite the series as ∞ n 1 e n=1 to recognize it as a geometric series with c = 1e and r = 1e . Thus, ∞ e−n = n=1 31. 1 e 1 − 1e = 1 . e−1 ∞ 8 + 2n ∞ 5en3−2n n=0 n=2 solution Rewrite the series as n ∞ ∞ n ∞ ∞ n 1 8 2 2 + = 8 · + , 5n 5n 5 5 n=0 n=0 n=0 n=0 0 which is a sum of two geometric series. The first series has c = 8 15 = 8 and r = 15 ; the second has c = 25 = 1 and r = 25 . Thus, n ∞ 1 8 8 8· = = 4 = 10, 1 5 1− 5 5 n=0 0 ∞ n 2 n=0 May 23, 2011 5 = 1 1 − 25 1 5 = 3 = , 3 5 S E C T I O N 10.2 Summing an Infinite Series 651 and ∞ 5 8 + 2n 35 = 10 + = . 5n 3 3 n=0 5 5 5 ∞+ 33. 5 − −n −3 5+n · · · 4 3(−2) 42 4 8n n=0 This is a geometric series with c = 5 and r = − 1 . Thus, solution 4 5 1 n 5 5 = 5· − = = 5 = 4. 1 1 4 1 + 1 − −4 4 4 n=0 ∞ 7 49 343 2401 −3 +4 2−5 26 + · · · 8 2 64+ 2 512 + 3 4096 + 4 + ··· 2 7 7 7 7 solution This is a geometric series with c = 78 and r = − 78 . Thus, 35. ∞ 7 7 7 n 7 7 8 = 8 = · − = . 15 7 8 8 15 1 − −8 8 n=0 37. Which of the following are not geometric series? ∞ 25 n+ 5 + 1 + 3 + 9 + 27 + · · · 97 3 5 25 125 (a) 29n (c) n=0 ∞ n=0 n2 2n (b) (d) ∞ 1 n4 n=3 ∞ π −n n=5 solution (a) ∞ n ∞ n 7 7 7 = : this is a geometric series with common ratio r = . 29n 29 29 n=0 n=0 (b) The ratio between two successive terms is 1 n4 an+1 (n+1)4 = = = 1 an (n + 1)4 4 n This ratio is not constant since it depends on n. Hence, the series 4 n . n+1 ∞ 1 is not a geometric series. n4 n=3 (c) The ratio between two successive terms is (n+1)2 an+1 n+1 = 2 2 n an n = 2 2n 1 2 1 (n + 1)2 · = 1 + · . n 2 n2 2n+1 This ratio is not constant since it depends on n. Hence, the series (d) ∞ n=5 π −n = ∞ n 1 n=5 π ∞ 2 n n=0 2n is not a geometric series. : this is a geometric series with common ratio r = 1 . π ∞ ∞ ∞ ∞ an converges and bn diverges, then (an + bn ) diverges. Hint: If not, derive a contradiction 39. Prove that if 1 diverges. Use the method of Example 8 to show that 1/3 n=1 n=1 n=1 k k=1 by writing ∞ n=1 May 23, 2011 bn = ∞ n=1 (an + bn ) − ∞ n=1 an 652 C H A P T E R 10 INFINITE SERIES ∞ ∞ solution Suppose to the contrary that ∞ n=1 an converges, n=1 bn diverges, but n=1 (an + bn ) converges. Then by the Linearity of Infinite Series, we have ∞ bn = n=1 so that ∞ (an + bn ) − n=1 ∞ an n=1 ∞ n=1 bn converges, a contradiction. ∞ tonshown that each of the following statements is false. Give a counterexample 9 +2 . Prove the divergence of ∞ 5n to zero, then an = 0. (a) If the general term an tends n=0 41. n=1 (b) The Nth partial sum of the infinite series defined by {an } is aN . ∞ (c) If an tends to zero, then an converges. n=1 ∞ (d) If an tends to L, then an = L. n=1 solution (a) Let an = 2−n . Then limn→∞ an = 0, but an is a geometric series with c = 20 = 1 and r = 1/2, so its sum is 1 = 2. 1 − (1/2) (b) Let an = 1. Then the nth partial sum is a1 + a2 + · · · + an = n while an = 1. ∞ 1 (c) Let an = √ . An example in the text shows that while an tends to zero, the sum an does not converge. n=1 n ∞ (d) Let an = 1. Then clearly an tends to L = 1, while the series n=1 an obviously diverges. 43. Compute the total area of the (infinitely many) triangles in Figure 4. ∞ 2 Suppose that S = an is an infinite series with partial sum SN = 5 − 2 . N y n=1 10 161 (a) What are the values of an and 2 an ? n=5 n=1 (b) What is the value of a3 ? (c) Find a general formula for an . ∞ (d) Find the sum an . 1 1 16 8 1 4 x 1 2 1 FIGURE 4 n=1 solution The area of a triangle with base B and height H is A = 12 BH . Because all of the triangles in Figure 4 have height 12 , the area of each triangle equals one-quarter of the base. Now, for n ≥ 0, the nth triangle has a base which 1 to x = 1 . Thus, extends from x = n+1 2n 2 1 1 1 B = n − n+1 = n+1 2 2 2 and A= 1 1 B = n+3 . 4 2 The total area of the triangles is then given by the geometric series ∞ n=0 1 = 2n+3 ∞ 1 1 1 1 n = 8 1 = . 8 2 4 1− 2 n=0 45. Find the total length of the infinite zigzag path in Figure 5 (each zag occurs at an angle of π4 ). The winner of a lottery receives m dollars at the end of each year for N years. The present value (PV) of this prize N in today’s dollars is PV = m(1 + r)−i , where r is the interest rate. Calculate PV if m = $50,000, r = 0.06, and i=1 N = 20. What is PV if N = ∞? π /4 π /4 1 FIGURE 5 May 23, 2011 S E C T I O N 10.2 Summing an Infinite Series 653 solution Because the angle at the lower left in Figure 5 has measure π4 and each zag in the path occurs at an angle of π , every triangle in the figure is an isosceles right triangle. Accordingly, the length of each new segment in the path is 4 √1 times the length of the previous segment. Since the first segment has length 1, the total length of the path is 2 √ ∞ √ 1 2 1 n = = √ = 2 + 2. √ 1 √ 2 2−1 1− n=0 2 ∞ 1 integer, then 47. Show that if a is a positive Evaluate . Hint: Find constants A, B, and C such that n(n + 1)(n + 2) ∞ n=1 1 1 1C 11 = = 1A++ +B· · · + + a)+ 2)a n 2 n + 1 + an + 2 n(nn(n + 1)(n n=1 solution By partial fraction decomposition 1 A B = + ; n (n + a) n n+a clearing the denominators gives 1 = A(n + a) + Bn. Setting n = 0 then yields A = a1 , while setting n = −a yields B = − a1 . Thus, 1 1 1 1 = a − a = n (n + a) n n+a a 1 1 − n n+a , and ∞ n=1 ∞ 1 1 1 1 = − . n(n + a) a n n+a n=1 For N > a, the Nth partial sum is 1 1 1 1 1 1 1 1 1 1 + + + ··· + − + + + ··· + . SN = a 2 3 a a N +1 N +2 N +3 N +a Thus, 1 1 1 1 1 = lim SN = 1 + + + ··· + . n(n + a) N→∞ a 2 3 a n=1 ∞ 49. LetA{bball a sequence and let a = b − b . Show that andground, only ifit lim bn to exists. n } be n n n converges n−1 dropped from a height of 10 ft begins to bounce. Eachatime it strikesifthe returns two-thirds of n→∞ its previous height. What is the total distance traveled by then=1 ball if it bounces infinitely many times? ∞ solution Let an = bn − bn−1 . The general term in the sequence of partial sums for the series an is then ∞ n=1 SN = (b1 − b0 ) + (b2 − b1 ) + (b3 − b2 ) + · · · + (bN − bN−1 ) = bN − b0 . Now, if lim bN exists, then so does lim SN and N→∞ N→∞ ∞ an converges. On the other hand, if n=1 lim SN exists, which implies that lim bN also exists. Thus, N→∞ N→∞ ∞ ∞ an converges, then n=1 an converges if and only if lim bn exists. n→∞ n=1 Assumptions Show, by giving counterexamples, that the assertions of Theorem 1 are not valid if the Further Insights Matter and Challenges ∞ ∞ n=0 n=0 use the Exercises 51–53 formula an and bn are not convergent. series 1 + r + r 2 + · · · + r N−1 = 1 − rN 1−r 7 51. Professor GeorgeAndrews of Pennsylvania State University observed that we can use Eq. (7) to calculate the derivative of f (x) = x N (for N ≥ 0). Assume that a = 0 and let x = ra. Show that f (a) = lim x→a and evaluate the limit. May 23, 2011 x N − aN rN − 1 = a N−1 lim x−a r→1 r − 1 654 C H A P T E R 10 INFINITE SERIES solution According to the definition of derivative of f (x) at x = a x N − aN . x→a x − a f (a) = lim Now, let x = ra. Then x → a if and only if r → 1, and N − aN N − aN aN r N − 1 x rN − 1 (ra) = lim = lim = a N−1 lim . f (a) = lim x→a x − a ra − a r→1 r→1 a (r − 1) r→1 r − 1 By Eq. (7) for a geometric sum, rN − 1 1 − rN = = 1 + r + r 2 + · · · + r N−1 , 1−r r −1 so rN − 1 = lim 1 + r + r 2 + · · · + r N−1 = 1 + 1 + 12 + · · · + 1N−1 = N. r→1 r − 1 r→1 lim Therefore, f (a) = a N−1 · N = N a N−1 53. Verify the Gregory–Leibniz formula as follows. de2 in Fermat used tothat compute the area under the graph of f (x) = x N over [0, A]. For (a) Set Pierre r = −x Eq. (7) andgeometric rearrange series to show 0 < r < 1, let F (r) be the sum of the areas of the infinitely many right-endpoint rectangles with endpoints Ar n , as in Figure 6. As r tends to 1,1 the rectangles become narrower and F (r) tends (−1) to theNarea x 2N under the graph. =1 − 1− x 2 + x 4 − · · · + (−1)N−1 x 2N−2 + r 2 1 N+1 +x 1 + x2 (a) Show that F (r) = A . 1 − r N+1 1], (b) Show, by integrating over [0, A that x N dx = lim F (r). (b) Use Eq. (7) to evaluate 1 2N 1 1 r→1 x dx 1 (−1)N−1 π 0 = 1 − + − + ··· + + (−1)N 4 3 5 7 2N − 1 0 1 + x2 (c) Use the Comparison Theorem for integrals to prove that 1 2N 1 x dx ≤ 0≤ 2N + 1 0 1 + x2 Hint: Observe that the integrand is ≤ x 2N . (d) Prove that 1 1 1 1 π = 1 − + − + − ··· 4 3 5 7 9 1 , and thereby conclude that Hint: Use (b) and (c) to show that the partial sums SN of satisfy SN − π4 ≤ 2N+1 π lim SN = 4 . N→∞ solution (a) Start with Eq. (7), and substitute −x 2 for r: 1 + r + r 2 + · · · + r N−1 = 1 − rN 1−r 1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 = 1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 = 1 − (−1)N x 2N 1 − (−x 2 ) 1 (−1)N x 2N − 1 + x2 1 + x2 (−1)N x 2N 1 = 1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 + 2 1+x 1 + x2 (b) The integrals of both sides must be equal. Now, 1 1 −1 x = tan−1 1 − tan−1 0 = π dx = tan 2 4 0 1+x 0 1 while 1 0 (−1)N x 2N 1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 + 1 + x2 May 23, 2011 dx S E C T I O N 10.2 Summing an Infinite Series 655 1 2N 1 1 1 x dx = x − x 3 + x 5 + · · · + (−1)N−1 x 2N−1 + (−1)N 3 5 2N − 1 0 1 + x2 1 2N 1 x dx 1 1 + (−1)N = 1 − + + · · · + (−1)N−1 3 5 2N − 1 0 1 + x2 (c) Note that for x ∈ [0, 1], we have 1 + x 2 ≥ 1, so that 0≤ x 2N ≤ x 2N 1 + x2 By the Comparison Theorem for integrals, we then see that 1 1 1 2N 1 1 x dx 2N 2N+1 = x ≤ x dx = 0≤ 2 2N + 1 2N +1 1 + x 0 0 0 (d) Write an = (−1)n 1 , 2n − 1 n≥1 and let SN be the partial sums. Then 1 2N 1 x 2N dx π 1 x dx N ≤ = SN − = (−1) 2 2 4 2N +1 1 + x 1 + x 0 0 Thus limN→∞ SN = π so that 4 π 1 1 1 1 = 1 − + − + − ... 4 3 5 7 9 55. The Koch snowflake (described in 1904 by Swedish mathematician Helge von Koch) is an infinitely jagged “fractal” Cantor’sasDisappearing Table (following Knop of but Hamilton Take table of length (Figure curve obtained a limit of polygonal curves (it Larry is continuous has noCollege) tangent line at aany point). BeginL with an 7). At stage 1, remove section of length centered at theedge midpoint. Twoedges sections remain, each with length less equilateral triangle (stage 0)the and produce stage 1L/4 by replacing each with four of one-third the length, arranged 2 L/2. stage 2,the remove sections length L/4 fromeach eachedge of these stage removes L/8 of the as inthan Figure 8. At Continue process: At theofnth stage, replace withtwo foursections edges of(this one-third the length. table). Now four sections remain, each of length less than L/4. At stage 3, remove the four central sections of length 4 (a) Show that the perimeter Pn of the polygon at the nth stage satisfies Pn = 3 Pn−1 . Prove that lim Pn = ∞. The n→∞ L/43 , etc. snowflake has infinite length. (a) Show that at the N th stage, each remaining section has length lessn−1 than L/2N and that the total amount of table new triangles are added at the nth stage, (b) Let A0 be the area of the original equilateral triangle. Show that (3)4 removed is 8 each with area A0 /9n (for n ≥ 1). Show that the is 5 A0 . total area of the Koch snowflake 1 1 1 1 + + + · · · + N+1 L 4 8 16 2 (b) Show that in the limit as N → ∞, precisely one-half of the table remains. This result is curious, because there are no nonzero intervals of table left (at each stage, the remaining sections have Stage 1 Stage 2 Stageplace 3 a length less than L/2N ). So the table has “disappeared.” However, we can any object longer than L/4 on the 8 of the removed sections. table. It will not fall through because it will not fitFIGURE through any solution (a) Each edge of the polygon at the (n − 1)st stage is replaced by four edges of one-third the length; hence the perimeter of the polygon at the nth stage is 43 times the perimeter of the polygon at the (n − 1)th stage. That is, Pn = 43 Pn−1 . Thus, P1 = 4 P0 ; 3 P2 = 4 P1 = 3 2 4 P0 , 3 n and, in general, Pn = 43 P0 . As n → ∞, it follows that P3 = 4 P2 = 3 3 4 P0 , 3 n 4 = ∞. n→∞ 3 lim Pn = P0 lim n→∞ (b) When each edge is replaced by four edges of one-third the length, one new triangle is created. At the (n − 1)st stage, there are 3 · 4n−1 edges in the snowflake, so 3 · 4n−1 new triangles are generated at the nth stage. Because the area of an equilateral triangle is proportional to the square of its side length and the side length for each new triangle is one-third the side length of triangles from the previous stage, it follows that the area of the triangles added at each stage is reduced by a factor of 19 from the area of the triangles added at the previous stage. Thus, each triangle added at the nth stage has an area of A0 /9n . This means that the nth stage contributes n A 3 4 3 · 4n−1 · n0 = A0 9 4 9 May 23, 2011 656 C H A P T E R 10 INFINITE SERIES to the area of the snowflake. The total area is therefore ∞ n 4 8 3 3 3 4 4 A = A0 + A0 = A0 + A0 9 4 = A0 + A0 · = A0 . 4 9 4 1− 4 5 5 9 n=1 10.3 Convergence of Series with Positive Terms Preliminary Questions 1. Let S = ∞ an . If the partial sums SN are increasing, then (choose the correct conclusion): n=1 (a) {an } is an increasing sequence. (b) {an } is a positive sequence. solution The correct response is (b). Recall that SN = a1 + a2 + a3 + · · · + aN ; thus, SN − SN−1 = aN . If SN is increasing, then SN − SN−1 ≥ 0. It then follows that aN ≥ 0; that is, {an } is a positive sequence. 2. What are the hypotheses of the Integral Test? solution The hypotheses for the Integral Test are: A function f (x) such that an = f (n) must be positive, decreasing, and continuous for x ≥ 1. ∞ n−3.2 converges? 3. Which test would you use to determine whether solution converges. Because n−3.2 = n=1 1 , we see that the indicated series is a p-series with p = 3.2 > 1. Therefore, the series n3.2 ∞ 1 4. Which test would you use to determine whether √ n 2 + n n=1 converges? solution Because 1 1 √ < n = 2 2n + n n 1 , 2 and ∞ n 1 n=1 2 is a convergent geometric series, the comparison test would be an appropriate choice to establish that the given series converges. 5. Ralph hopes to investigate the convergence of ∞ −n ∞ e 1 by comparing it with . Is Ralph on the right track? n n n=1 solution n=1 No, Ralph is not on the right track. For n ≥ 1, e−n 1 < ; n n ∞ 1 is a divergent series. The Comparison Test therefore does not allow us to draw a conclusion about the n n=1 ∞ −n e convergence or divergence of the series . n however, n=1 Exercises In Exercises 1–14, use the Integral Test to determine whether the infinite series is convergent. 1. ∞ 1 n4 n=1 1 Let f (x) = 4 . This function is continuous, positive and decreasing on the interval x ≥ 1, so the Integral x Test applies. Moreover, solution May 23, 2011 S E C T I O N 10.3 Convergence of Series with Positive Terms 657 ∞ R 1 1 dx −4 dx = − 1 lim = lim x − 1 = . 3 R→∞ R 3 3 R→∞ 1 x4 1 The integral converges; hence, the series ∞ 1 also converges. n4 n=1 3. ∞ ∞−1/3 n 1 n=1 n+3 n=1 1 1 Let f (x) = x − 3 = √ . This function is continuous, positive and decreasing on the interval x ≥ 1, so the 3 x Integral Test applies. Moreover, R ∞ 3 lim R 2/3 − 1 = ∞. x −1/3 dx = lim x −1/3 dx = 2 R→∞ R→∞ 1 1 solution The integral diverges; hence, the series ∞ n−1/3 also diverges. n=1 ∞ n2 ∞ 5. 1 3 + 9)5/2 (n√ n=25 n −4 n=5 solution Let f (x) = x2 5/2 . This function is positive and continuous for x ≥ 25. Moreover, because x3 + 9 f (x) = 2x(x 3 + 9) 5/2 3/2 − x 2 · 52 (x 3 + 9) · 3x 2 (x 3 + 9) 5 = x(36 − 11x 3 ) 2(x 3 + 9) 7/2 , we see that f (x) < 0 for x ≥ 25, so f is decreasing on the interval x ≥ 25. The Integral Test therefore applies. To evaluate the improper integral, we use the substitution u = x 3 + 9, du = 3x 2 dx. We then find ∞ R R 3 +9 1 x2 x2 du lim dx = lim dx = 3 R→∞ 15634 u5/2 R→∞ 25 (x 3 + 9)5/2 25 (x 3 + 9)5/2 2 1 1 2 = − lim − . = 9 R→∞ (R 3 + 9)3/2 156343/2 9 · 156343/2 The integral converges; hence, the series 7. ∞ n2 5/2 3 n=25 n + 9 also converges. ∞ ∞ 1 n n2 + 12 n=1 (n + 1)3/5 n=1 1 Let f (x) = 2 . This function is positive, decreasing and continuous on the interval x ≥ 1, hence the x +1 Integral Test applies. Moreover, R ∞ dx dx −1 R − π = π − π = π . tan = lim = lim 4 2 4 4 R→∞ 1 x 2 + 1 R→∞ 1 x2 + 1 solution The integral converges; hence, the series 9. ∞ 1 n=1 n2 + 1 also converges. ∞ ∞ 1 1 n(n + 1) n=1 n2 − 1 n=4 1 . This function is positive, continuous and decreasing on the interval x ≥ 1, so the x(x + 1) Integral Test applies. We compute the improper integral using partial fractions: R ∞ 1 1 1 x R 1 R dx = lim − dx = lim ln − ln = ln 1 − ln = ln 2. = lim ln x x+1 x + 1 1 R+1 2 2 R→∞ 1 R→∞ R→∞ 1 x(x + 1) solution Let f (x) = The integral converges; hence, the series ∞ n=1 May 23, 2011 1 converges. n(n + 1) 658 C H A P T E R 10 INFINITE SERIES 11. ∞ ∞ 1 2 −n 2 n(lnnen) n=2 n=1 solution Let f (x) = 1 x(ln x)2 f (x) = − . This function is positive and continuous for x ≥ 2. Moreover, 1 x 2 (ln x)4 1 · (ln x)2 + x · 2 (ln x) · 1 x =− 1 (ln x)2 + 2 ln x . x 2 (ln x)4 Since ln x > 0 for x > 1, f (x) is negative for x > 1; hence, f is decreasing for x ≥ 2. To compute the improper integral, 1 we make the substitution u = ln x, du = dx. We obtain: x ∞ R ln R 1 1 du dx = lim dx = lim 2 2 R→∞ R→∞ x(ln x) x(ln x) 2 2 ln 2 u2 1 1 1 − = . = − lim ln 2 ln 2 R→∞ ln R The integral converges; hence, the series ∞ n=2 1 also converges. n(ln n)2 ∞ ∞1 13. ln n 2ln n 2 n=1 n n=1 solution Note that 2ln n = (eln 2 )ln n = (eln n )ln 2 = nln 2 . Thus, ∞ 1 n=1 2ln n = ∞ n=1 1 . nln 2 1 . This function is positive, continuous and decreasing on the interval x ≥ 1; therefore, the Integral x ln 2 Test applies. Moreover, ∞ R dx dx 1 lim (R 1−ln 2 − 1) = ∞, = lim = 1 − ln 2 R→∞ R→∞ 1 x ln 2 1 x ln 2 Now, let f (x) = because 1 − ln 2 > 0. The integral diverges; hence, the series ∞ n=1 1 also diverges. 2ln n ∞ ∞ 1 n−3 . converges by using the Comparison Test with 3 + 8n n n n=1 3lnn=1 ∞ n=1 solution We compare the series with the p-series n−3 . For n ≥ 1, ∞ 15. Show that1 n=1 1 1 ≤ 3. n3 + 8n n Since ∞ ∞ 1 1 converges (it is a p-series with p = 3 > 1), the series also converges by the Comparison Test. n3 n3 + 8n n=1 n=1 ∞ ∞1 ∞ 17. Let S = √ . 1Verify that for n ≥ 1, n−1 . diverges by comparing with Show thatn + n n=1 2−3 n n=2 n=2 1 1 1 1 √ ≤ , √ ≤ √ n n+ n n+ n n 1 1 Can either inequality be used to show that S diverges? Show that and conclude that S diverges. √ ≥ 2n n+ n √ √ √ solution For n ≥ 1, n + n ≥ n and n + n ≥ n. Taking the reciprocal of each of these inequalities yields 1 1 √ ≤ n n+ n May 23, 2011 and 1 1 √ ≤ √ . n n+ n S E C T I O N 10.3 These inequalities indicate that the series ∞ n=1 Convergence of Series with Positive Terms 659 ∞ ∞ ∞ 1 1 1 1 and and √ is smaller than both √ ; however, n n n+ n n ∞ 1 √ both diverge so neither inequality allows us to show that S diverges. n n=1 √ √ On the other hand, for n ≥ 1, n ≥ n, so 2n ≥ n + n and n=1 n=1 n=1 1 1 . √ ≥ 2n n+ n The series ∞ ∞ 1 1 =2 diverges, since the harmonic series diverges. The Comparison Test then lets us conclude 2n n n=1 n=1 that the larger series ∞ n=1 1 √ also diverges. n+ n In Exercises 19–30, use the Comparison Test to determine whether the infinite series is ∞convergent. 1 Which of the following inequalities can be used to study the convergence of √ ? Explain. ∞ 2 1 n + n n=2 19. n2n n=1 1 1 1 1 ,n ∞√ √ ≤ √ ≤ 2 2 + n n1 2+ n n n n solution We compare with the geometric series . For n ≥ 1, 2 n=1 1 1 ≤ n = n2n 2 Since ∞ n 1 2 n=1 n 1 . 2 converges (it is a geometric series with r = 12 ), we conclude by the Comparison Test that ∞ 1 also n2n n=1 converges. ∞ ∞ 1 n3 n1/3 5+ 2n n=1 n + 4n + 1 n=1 solution For n ≥ 1, 21. 1 1 ≤ n 2 n1/3 + 2n The series ∞ 1 1 1 , so it converges. By the Comparison test, so does is a geometric series with r = . n=1 2n 2 n1/3 + 2n ∞ n=1 ∞ ∞ 4 23. 1 m! + 4m m=1 3 n=1 n + 2n − 1 solution For m ≥ 1, 4 4 ≤ m = m m! + 4 4 The series m−1 1 . 4 ∞ m−1 1 1 is a geometric series with r = , so it converges. By the Comparison Test we can therefore 4 4 m=1 conclude that the series ∞ m=1 4 also converges. m! + 4m ∞ sin2 √ k ∞ 25. n k2 k=1 n−3 n=4 solution For k ≥ 1, 0 ≤ sin2 k ≤ 1, so 0≤ The series 1 sin2 k ≤ 2. 2 k k ∞ 1 is a p-series with p = 2 > 1, so it converges. By the Comparison Test we can therefore conclude that k2 k=1 ∞ sin2 k also converges. the series k2 k=1 May 23, 2011 660 C H A P T E R 10 INFINITE SERIES 27. ∞ ∞ 2 1/3 k 3n + 3−n 5/4 n=1 k −k k=2 Since 3−n > 0 for all n, solution n 2 2 1 ≤ = 2 . 3n + 3−n 3n 3 The series ∞ n 1 1 2 is a geometric series with r = , so it converges. By the Comparison Theorem we can therefore 3 3 n=1 conclude that the series ∞ n=1 29. 2 also converges. 3n + 3−n ∞ ∞ 1 2 (n + 1)! 2−k n=1 k=1 Note that for n ≥ 2, solution (n + 1)! = 1 · 2 · 3 · · · n · (n + 1) ≤ 2n n factors so that ∞ n=1 But ∞ ∞ n=2 n=2 1 1 1 =1+ ≤1+ (n + 1)! (n + 1)! 2n ∞ 1 1 1 converges as n=2 2n is a geometric series with ratio r = 2 , so it converges. By the comparison test, (n + 1)! ∞ n=1 well. Exercise 31–36: For all a > 0 and b > 1, the inequalities ∞ n! n=1 ln n ≤ na , n3 na < bn are true for n sufficiently large (this can be proved using L’Hopital’s Rule). Use this, together with the Comparison Theorem, to determine whether the series converges or diverges. 31. ∞ ln n n3 n=1 solution For n sufficiently large (say n = k, although in this case n = 1 suffices), we have ln n ≤ n, so that ∞ ∞ ∞ ln n n 1 ≤ = 3 3 n n n2 n=k n=k This is a p-series with p = 2 > 1, so it converges. Thus terms for 1 ≤ n ≤ k does not affect this result. 33. n=k ∞ ln n n=k n3 also converges; adding back in the finite number of ∞ (ln n)100 ∞ 1 n1.1 n=1 ln m m=2 solution Choose N so that ln n ≤ n0.0005 for n ≥ N . Then also for n > N, (ln n)100 ≤ (n0.0005 )100 = n0.05 . Then ∞ ∞ ∞ (ln n)100 n0.05 1 ≤ = 1.1 1.1 1.05 n n n n=N But ∞ n=N n=N n=N 1 (ln n)1 00 is a p-series with p = 1.05 > 1, so is convergent. It follows that ∞ is also convergent; n=N n1.1 n1.05 adding back in the finite number of terms for n = 1, 2, . . . , N − 1 shows that ∞ (ln n)100 converges as well. n1.1 n=1 ∞ n=1 1 (ln n)10 May 23, 2011 Convergence of Series with Positive Terms S E C T I O N 10.3 35. 661 ∞ n 3n n=1 Choose N such that n ≤ 2n for n ≥ N . Then solution ∞ ∞ n n 2 ≤ n 3 3 n=N n=N 2 < 1, so it converges. Thus the series on the left converges as well. Adding 3 ∞ n converges. back in the finite number of terms for n < N shows that 3n The latter sum is a geometric series with r = n=1 ∞ 1 sin 2 converges. Hint: Use sin x ≤ x for x ≥ 0. n ∞ 5 37. Show that n 2n n=1 n=1 solution For n ≥ 1, 1 0 ≤ 2 ≤ 1 < π; n therefore, sin 12 > 0 for n ≥ 1. Moreover, for n ≥ 1, n 1 1 sin 2 ≤ 2 . n n ∞ 1 is a p-series with p = 2 > 1, so it converges. By the Comparison Test we can therefore conclude that n2 The series n=1 ∞ the series n=1 1 sin 2 also converges. n In Exercises 39–48, use the Limit Comparison Test to prove convergence or divergence of the infinite series. ∞ sin(1/n) Does converge? ∞ 2 n ln n n=2 39. n4 − 1 n=2 n2 1 1 n2 n2 Let an = 4 . For large n, 4 ≈ 4 = 2 , so we apply the Limit Comparison Test with bn = 2 . n −1 n −1 n n n solution We find n2 4 an n4 L = lim = lim n 1−1 = lim 4 = 1. n→∞ bn n→∞ n→∞ n − 1 2 n The series ∞ n=1 ∞ 1 1 is a p-series with p = 2 > 1, so it converges; hence, also converges. Because L exists, by the n2 n2 Limit Comparison Test we can conclude that the series ∞ n=2 n=2 n2 converges. 4 n −1 ∞ ∞ n 41. 1 3 √ n=2 nn2+−1 n n=2 solution n 1 n n Let an = . For large n, ≈ √ = √ , so we apply the Limit Comparison test with 3 3 3 n n n +1 n +1 1 bn = √ . We find n an = lim L = lim n→∞ bn n→∞ The series √ √n n3 n3 +1 = lim = 1. 1 n→∞ √ n3 + 1 n ∞ ∞ 1 1 √ is a p-series with p = 12 < 1, so it diverges; hence, √ also diverges. Because L > 0, by the n n n=1 Limit Comparison Test we can conclude that the series ∞ n=2 May 23, 2011 n=2 n diverges. n3 + 1 662 C H A P T E R 10 INFINITE SERIES 43. ∞ ∞ 3n + 53 n n(n− 1)(n − 2) n=3 7 2 n=2 n + 2n + 1 Let an = solution 3n + 5 3 3n + 5 3n . For large n, ≈ 3 = 2 , so we apply the Limit Comparison n(n − 1)(n − 2) n(n − 1)(n − 2) n n 1 Test with bn = 2 . We find n 3n+5 an 3n3 + 5n2 n(n+1)(n+2) = lim = lim = 3. 1 n→∞ bn n→∞ n→∞ n(n + 1)(n + 2) 2 L = lim n The series ∞ n=1 ∞ 1 1 is a p-series with p = 2 > 1, so it converges; hence, the series also converges. Because L n2 n2 exists, by the Limit Comparison Test we can conclude that the series ∞ n=3 n=3 3n + 5 converges. n(n − 1)(n − 2) ∞ ∞ 1n 45. √ e +n n+ ln n n=1 e2n − n2 n=1 solution Let 1 an = √ n + ln n For large n, √ n + ln n ≈ √ 1 n, so apply the Comparison Test with bn = √ . We find n √ an 1 n 1 = lim = lim √ =1 · L = lim n→∞ bn n→∞ n + ln n n→∞ 1 √n 1 + ln n The series ∞ 1 1 √ is a p-series with p = < 1, so it diverges. Because L exists, the Limit Comparison Test tells us the 2 n n=1 the original series also diverges. ∞ ∞ 1 ∞ 47. n−2 . 1− cos ln(n + 4) Hint: Compare with n 5/2 n=1 n=1 n n=1 1 1 solution Let an = 1 − cos , and apply the Limit Comparison Test with bn = 2 . We find n n − 12 sin x1 1 − cos n1 1 − cos x1 sin x1 1 an x lim = lim = lim = lim = . 1 1 n→∞ bn n→∞ x→∞ x→∞ 2 x→∞ 1 − 23 2 2 L = lim n x x x As x → ∞, u = x1 → 0, so L= sin x1 1 1 1 sin u lim = . = lim 2 x→∞ 1 2 u→0 u 2 x ∞ 1 is a p-series with p = 2 > 1, so it converges. Because L exists, by the Limit Comparison Test we can n2 n=1 ∞ 1 also converges. 1 − cos conclude that the series n The series n=1 In Exercises ∞ 49–78, determine convergence or divergence using any method covered so far. (1 − 2−1/n ) Hint: Compare with the harmonic series. ∞ 1 n=1 49. n2 − 9 n=4 1 1 solution Apply the Limit Comparison Test with an = 2 and bn = 2 : n −9 n 1 2 an n2 = lim n 1−9 = lim 2 = 1. n→∞ bn n→∞ n→∞ n − 9 2 L = lim n May 23, 2011 S E C T I O N 10.3 Since the p-series Convergence of Series with Positive Terms 663 ∞ ∞ 1 1 converges, the series also converges. Because L exists, by the Limit Comparison Test n2 n2 n=1 we can conclude that the series √ ∞ ∞ n 2 51. cos n 4n + 9 n=1 n2 ∞ n=4 n=4 1 converges. 2 n −9 √ n=1 solution Apply the Limit Comparison Test with an = 1 n and bn = √ : 4n + 9 n √ n 1 an n = . = lim 4n+9 = lim L = lim n→∞ bn n→∞ √1 n→∞ 4n + 9 4 n ∞ 1 The series √ is a divergent p-series. Because L > 0, by the Limit Comparison Test we can conclude that the series n n=1 √ ∞ n also diverges. 4n + 9 n=1 53. ∞ 2 n −n ∞ n − cos n n5 + n 3 n=1 n n=1 solution n (n − 1) n2 − n n−1 = = 4 First rewrite an = 5 and observe n +n n n +1 n4 + 1 1 n n−1 < 4 = 3 n4 + 1 n n for n ≥ 1. The series ∞ 1 is a convergent p-series, so by the Comparison Test we can conclude that the series n3 n=1 ∞ 2 n −n also converges. n5 + n n=1 55. ∞ ∞ (4/5)−n 1 n=5 n2 + sin n n=1 solution ∞ −n 4 n=5 5 = ∞ n 5 n=5 which is a geometric series starting at n = 5 with ratio r = ∞ ∞ 1 1 n3/2 ln2 n n=2 n n=1 3 solution For n ≥ 3, ln n > 1, so n3/2 ln n > n3/2 and 4 5 > 1. Thus the series diverges. 4 57. 1 n3/2 ln n The series 1 < 3/2 . n ∞ ∞ 1 1 is a convergent p-series, so the series also converges. By the Comparison Test we can 3/2 n n3/2 n=1 therefore conclude that the series ∞ n=3 n=3 ∞ 1 1 converges. Hence, the series also converges. 3/2 3/2 n ln n n ln n n=2 ∞ ∞1/k 59. 4 (ln n)12 k=1 n=2 n9/8 solution lim ak = lim 41/k = 40 = 1 = 0; k→∞ therefore, the series ∞ k→∞ 41/k diverges by the Divergence Test. k=1 May 23, 2011 664 C H A P T E R 10 INFINITE SERIES 61. ∞ ∞ 1 4n (ln n)n4 n=2 5 − 2n n=1 solution By the comment preceding Exercise 31, we can choose N so that for n ≥ N, we have ln n < n1/8 , so that (ln n)4 < n1/2 . Then ∞ n=N ∞ 1 1 > (ln n)4 n1/2 n=N which is a divergent p-series. Thus the series on the left diverges as well, and adding back in the finite number of terms ∞ 1 diverges. for n < N does not affect the result. Thus (ln n)4 n=2 63. ∞ ∞ 1 n 2 n ln nn − n n=1 3 −n n=1 solution For n ≥ 2, n ln n − n ≤ n ln n; therefore, 1 1 ≥ . n ln n − n n ln n 1 . For x ≥ 2, this function is continuous, positive and decreasing, so the Integral Test applies. Using x ln x the substitution u = ln x, du = x1 dx, we find Now, let f (x) = ∞ R ln R dx dx du = lim = lim = lim (ln(ln R) − ln(ln 2)) = ∞. x ln x x ln x u R→∞ 2 R→∞ ln 2 R→∞ 2 The integral diverges; hence, the series the series ∞ n=2 65. ∞ n=2 1 also diverges. By the Comparison Test we can therefore conclude that n ln n 1 diverges. n ln n − n ∞ ∞1 nn 1 n=1 n(ln n)2 − n n=1 For n ≥ 2, nn ≥ 2n ; therefore, solution 1 1 ≤ n = nn 2 The series ∞ n 1 n=1 2 ∞ n 1 is a convergent geometric series, so therefore conclude that the series n 1 . 2 n=2 2 ∞ ∞ 1 1 converges. Hence, the series converges. nn nn n=2 67. also converges. By the Comparison Test we can n=1 ∞ n 1 + (−1) ∞ n2 − 4n3/2 n n=1 n3 n=1 solution Let an = 1 + (−1)n n Then an = 0 2 1 2k = k n odd n = 2k even Therefore, {an } consists of 0s in the odd places and the harmonic series in the even places, so the harmonic series, which diverges. Thus ∞ i=1 an diverges as well. ∞ 2 + (−1)n n3/2 n=1 May 23, 2011 ∞ i=1 an is just the sum of Convergence of Series with Positive Terms S E C T I O N 10.3 69. ∞ sin n=1 665 1 n solution Apply the Limit Comparison Test with an = sin L = lim sin n1 n→∞ 1 n 1 1 and bn = : n n = lim u→0 sin u = 1, u where u = n1 . The harmonic series diverges. Because L > 0, by the Limit Comparison Test we can conclude that the ∞ 1 sin also diverges. series n n=1 71. ∞ 2n + 1 ∞ sin(1/n) 4n √ n=1 n n=1 solution For n ≥ 3, 2n + 1 < 2n , so 2n + 1 2n < n = n 4 4 The series ∞ n 1 n=1 2 is a convergent geometric series, so ∞ n 1 n=3 therefore conclude that the series n 1 . 2 2 also converges. By the Comparison Test we can ∞ ∞ 2n + 1 2n + 1 converges. Finally, the series converges. 4n 4n n=3 73. n=1 ∞ ∞ ln n 1 n2 −√3n n=4 n e n=3 solution By the comment preceding Exercise 31, we can choose N ≥ 4 so that for n ≥ N , ln n < n1/2 . Then ∞ n=N ∞ ∞ ln n n1/2 1 ≤ = 2 2 3/2 n − 3n n − 3n n − 3n1/2 n=N To evaluate convergence of the latter series, let an = Test: n=N 1 1 and bn = 3/2 , and apply the Limit Comparison n3/2 − 3n1/2 n an 1 1 = lim 3/2 · n3/2 = lim =0 1/2 n→∞ bn n→∞ n n→∞ − 3n 1 − 3n−1 L = lim Thus an converges if bn does. But bn is a convergent p-series. Thus an converges and, by the comparison test, so does the original series. Adding back in the finite number of terms for n < N does not affect convergence. 75. ∞ ∞ 1 1 n1/2 lnlnnn n=2 3 n=1 solution By the comment preceding Exercise 31, we can choose N ≥ 2 so that for n ≥ N, ln n < n1/4 . Then ∞ n=N ∞ 1 1 > 1/2 3/4 n ln n n n=N which is a divergent p-series. Thus the original series diverges as well - as usual, adding back in the finite number of terms for n < N does not affect convergence. ∞ 2 ∞ 4n + 15n 1 3n4 − 5n2 −417 n=1 3/2 − ln n n n=1 solution Apply the Limit Comparison Test with 77. an = May 23, 2011 4n2 + 15n , 3n4 − 5n2 − 17 bn = 4n2 4 = 2 3n4 3n 666 C H A P T E R 10 INFINITE SERIES We have an 4n2 + 15n 12n4 + 45n3 12 + 45/n 3n2 = lim =1 = lim · = lim n→∞ bn n→∞ 3n4 − 5n2 − 17 n→∞ 12n4 − 20n2 − 68 n→∞ 12 − 20/n2 − 68/n4 4 L = lim Now, ∞ n=1 bn is a p-series with p = 2 > 1, so converges. Since L = 1, we see that ∞ 4n2 + 15n n=1 3n4 − 5n2 − 17 converges as well. ∞ 1 ∞ 79. For which a does converge? n n(ln n)a 4−n + 5−nn=2 n=1 First consider the case a > 0 but a = 1. Let f (x) = solution 1 . This function is continuous, positive and x(ln x)a decreasing for x ≥ 2, so the Integral Test applies. Now, ∞ R ln R 1 1 1 dx dx du = lim = lim = − . lim a 1 − a R→∞ (ln R)a−1 R→∞ 2 x(ln x)a R→∞ ln 2 ua (ln 2)a−1 2 x(ln x) Because 1 ∞, 0 < a < 1 = lim R→∞ (ln R)a−1 0, a>1 we conclude the integral diverges when 0 < a < 1 and converges when a > 1. Therefore ∞ n=2 1 converges for a > 1 and diverges for 0 < a < 1. n(ln n)a Next, consider the case a = 1. The series becomes ∞ n=2 1 1 . Let f (x) = . For x ≥ 2, this function is continuous, n ln n x ln x positive and decreasing, so the Integral Test applies. Using the substitution u = ln x, du = x1 dx, we find R ln R ∞ dx dx du = lim = lim = lim (ln(ln R) − ln(ln 2)) = ∞. u R→∞ 2 x ln x R→∞ ln 2 R→∞ 2 x ln x The integral diverges; hence, the series also diverges. ∞ (ln n)b Finally, consider the case a < 0. Let b = −a > 0 so the series becomes n=2 n . Since ln n > 1 for all n ≥ 3, it follows that (ln n)b > 1 The series ∞ (ln n)b n=2 n ∞ 1 n=3 n so 1 (ln n)b > . n n diverges, so by the Comparison Test we can conclude that ∞ (ln n)b n=3 n also diverges. Consequently, diverges. Thus, ∞ n=2 1 diverges for a < 0. n(ln n)a To summarize: ∞ n=2 1 converges if a > 1 and diverges if a ≤ 1. n(ln n)a Approximating Infinite Sums ∞ In Exercises 81–83, let an = f (n), where f (x) is a continuous, decreasing function such ∞ 1 f (x) dx aconverges. that f (x) 0 and a 1does converge? For≥which n ln n n=2 81. Show that ∞ ∞ ∞ f (x) dx ≤ an ≤ a1 + f (x) dx 3 1 May 23, 2011 n=1 1 Convergence of Series with Positive Terms S E C T I O N 10.3 solution 667 From the proof of the Integral Test, we know that a2 + a3 + a4 + · · · + aN ≤ that is, SN − a1 ≤ ∞ N 1 f (x) dx ≤ or SN ≤ a1 + f (x) dx 1 ∞ f (x) dx; 1 ∞ f (x) dx. 1 Also from the proof of the Integral test, we know that N f (x) dx ≤ a1 + a2 + a3 + · · · + aN−1 = SN − aN ≤ SN . 1 Thus, N 1 f (x) dx ≤ SN ≤ a1 + ∞ f (x) dx. 1 Taking the limit as N → ∞ yields Eq. (3), as desired. ∞ Eq. showasthat 83. Let S Using = an . (3), Arguing in Exercise 81, show that n=1 M an + ∞ n=1 M+1 ∞ 1 5≤ M+1 ≤ 6 ∞ n1.2 f (x) dx ≤n=1 S≤ an + f (x) dx 4 M+1 n=1 This series converges slowly. Use a computer algebra system to verify that SN < 5 for N ≤ 43,128 and S43,129 ≈ Conclude that 5.00000021. ⎛ ⎞ ∞ M 0≤S−⎝ an + f (x) dx ⎠ ≤ aM+1 5 M+1 n=1 This provides a method for approximating S with an error of at most aM+1 . solution Following the proof of the Integral Test and the argument in Exercise 81, but starting with n = M + 1 rather than n = 1, we obtain ∞ M+1 Adding M ∞ f (x) dx ≤ an ≤ aM+1 + ∞ f (x) dx. M+1 n=M+1 an to each part of this inequality yields n=1 M n=1 Subtracting M n=1 an + ∞ M+1 an + ∞ M+1 f (x) dx ≤ an = S ≤ n=1 M+1 an + n=1 ∞ f (x) dx. M+1 f (x) dx from each part of this last inequality then gives us ⎛ 0≤S−⎝ M n=1 85. ∞ an + ∞ M+1 ⎞ f (x) dx ⎠ ≤ aM+1 . Apply Eq. (4) with M = 40,000 to show that Use Eq. (4) with M = 43,129 to prove that ∞ ∞ 1 1.644934066 ≤ 1≤ 1.644934068 n2 ≤ 5.5915839 5.5915810 ≤ n=1 n1.2 n=1 Is this consistent with Euler’s result, according to which this infinite series has sum π 2 /6? 1 1 solution Using Eq. (4) with f (x) = 2 , an = 2 and M = 40,000, we find x n ∞ ∞ ∞ dx 1 dx S40,000 + ≤ ≤ S + . 40,001 n2 40,001 x 2 40,001 x 2 n=1 May 23, 2011 668 C H A P T E R 10 INFINITE SERIES Now, S40,000 = 1.6449090672; S40,001 = S40,000 + 1 = 1.6449090678; 40,001 and R dx dx 1 1 1 = lim = − lim − = = 0.0000249994. 2 2 R 40,001 40,001 R→∞ 40,001 x R→∞ 40,001 x ∞ Thus, 1.6449090672 + 0.0000249994 ≤ ∞ 1 ≤ 1.6449090678 + 0.0000249994, n2 n=1 or 1.6449340665 ≤ ∞ 1 ≤ 1.6449340672. n2 n=1 π2 ≈ 1.6449340668, our approximation is consistent with Euler’s result. 6 ∞ 87. Using a CAS and Eq. (5), determine the value of∞ n−5 to within an error less than 10−4 . Using a CAS and Eq. (5), determine the value −5 of n=1 n−6 to within an error less than 10−4 . Check that your −5 solution Using Eq. (5) with f (x) = x and an = n , we have Since n=1 ⎛ proved that the sum is equal ⎞ to π 6 /945. result is consistent with that of∞Euler, who ∞ M+1 0≤ n−5 − ⎝ n−5 + x −5 dx ⎠ ≤ (M + 1)−5 . n=1 n=1 M+1 To guarantee an error less than 10−4 , we need (M + 1)−5 ≤ 10−4 . This yields M ≥ 104/5 − 1 ≈ 5.3, so we choose M = 6. Now, 7 n−5 = 1.0368498887, n=1 and ∞ 7 x −5 dx = lim R R→∞ 7 x −5 dx = − 1 1 lim R −4 − 7−4 = = 0.0001041233. 4 R→∞ 4 · 74 Thus, ∞ n=1 n−5 ≈ 7 n=1 n−5 + ∞ 7 x −5 dx = 1.0368498887 + 0.0001041233 = 1.0369540120. ∞ far canargument a stack of identical books (of mass and unit length) without tipping over? stackTest. will 89. TheHow following proves the divergence of themharmonic series Sextend = 1/n without using theThe Integral not tip over if the (n + 1)st book is placed at the bottom of the stack with itsn=1 right edge located at the center of mass Let of the first n books (Figure 5). Let cn be the center of mass of the first n books, measured along the x-axis, where we take the positive x-axis to the left of the origin as in Figure 6. Recall that if an object of mass m1 has center of mass 1 1 1 1 1 at x1 and a second object of m has1 + center+of mass + · · ·x,2 , thenSthe + of+mass+of · · the · system has x-coordinate S12 = 2 =center 3 5 2 4 6 m1 x1 + m2 x2 Show that if S converges, then m1 + m2 (a) S1 and S2 also converge and S = S1 + S2 . Show that if the1 (n + 1)st book is placed with its right edge at cn , then its center of mass is located at cn + 12 . (b) S(a) 1 > S2 and S2 = 2 S. (b) Consider the first n books as aconclude single object mass nm with center of mass at cn and the (n + 1)st book as a second Observe that (b) contradicts (a), and that Sofdiverges. 1 . object ofAssume mass m.throughout Show that if theS (n + 1)st book placed with its right edge Write at cn , then cn+1 = cn + solution that converges; we is will derive a contradiction. 2(n + 1) (c) Prove that lim cn = ∞. Thus, by using enough books, the stack can be extended as far as desired without n→∞ 1 1 1 an = , bn = , cn = tipping over. n 2n − 1 2n May 23, 2011 Convergence of Series with Positive Terms S E C T I O N 10.3 669 for the nth terms in the series S, S1 , and S2 . Since 2n − 1 ≥ n for n ≥ 1, we have bn < an . Since S = an converges, 1 so does S1 = bn by the Comparison Test. Also, cn = an , so again by the Comparison Test, the convergence of S 2 implies the convergence of S2 = cn . Now, define two sequences b n odd bn = (n+1)/2 0 n even 0 n odd cn = cn/2 n even and cn look bn and cn , but have inserted in the “missing” places compared to an . Then an = bn + cn ; That is, bn like zeros also S1 = bn = bn and S2 = cn = cn . Finally, since S, S1 , and S2 all converge, we have S= ∞ an = n=1 ∞ ∞ (bn + cn ) = n=1 bn + n=1 ∞ cn = n=1 ∞ bn + n=1 ∞ cn = S1 + S2 n=1 1 Now, bn > cn for every n, so that S1 > S2 . Also, we showed above that cn = an , so that 2S2 = S. Putting all this 2 together gives S = S1 + S2 > S2 + S2 = 2S2 = S so that S > S, a contradiction. Thus S must diverge. Further Insights and Challenges ∞ ∞ 91. Kummer’s Acceleration Method Suppose we wish to approximate S = 1/n2 . There is a similar telescoping − ln n Let S = an , where an = (ln(ln n)) . n=1 n=2can be computed exactly (Example 1 in Section 10.2): series whose value n)) . (a) Show, by taking logarithms, that an = n− ln(ln(ln ∞ 12 ee . = 1 (b) Show that ln(ln(ln n)) ≥ 2 if n > C, where C = n(ne + 1) n=1 (c) Show that S converges. (a) Verify that S= ∞ n=1 ∞ 1 + n(n + 1) n=1 1 1 − n(n + 1) n2 Thus for M large, S ≈1+ M n=1 1 n2 (n + 1) 6 (b) Explain what has been gained. Why is Eq. (6) a better approximation to S than is M 1/n2 ? n=1 Compute (c) 1000 n=1 1 , n2 1+ 100 n=1 1 n2 (n + 1) Which is a better approximation to S, whose exact value is π 2 /6? solution (a) Because the series ∞ ∞ 1 1 and both converge, n(n + 1) n2 n=1 ∞ n=1 1 + n(n + 1) n=1 ∞ n=1 1 1 − 2 n(n + 1) n = ∞ n=1 ∞ ∞ ∞ 1 1 1 1 + = − = S. 2 n(n + 1) n(n + 1) n n2 n=1 n=1 Now, n+1 1 1 n 1 = 2 − − = 2 , n(n + 1) n2 n (n + 1) n2 (n + 1) n (n + 1) May 23, 2011 n=1 670 C H A P T E R 10 INFINITE SERIES so, for M large, S ≈1+ M n=1 (b) The series ∞ 1 n=1 n2 (n+1) converges more rapidly than 1 . n2 (n + 1) ∞ 1 since the degree of n in the denominator is larger. n2 n=1 (c) Using a computer algebra system, we find 1000 n=1 1 = 1.6439345667 n2 and 1+ 100 n=1 1 = 1.6448848903. n2 (n + 1) The second sum is more accurate because it is closer to the exact solution The series S = ∞ π2 ≈ 1.6449340668. 6 k −3 has been computed to more than 100 million digits. The first 30 digits are 10.4 Absolute andk=1Conditional Convergence S = 1.202056903159594285399738161511 S using Acceleration of Exercise = 100 and auxiliary series 1. Approximate Give an example of a the series such that Method an converges but 91 with |an |M diverges. (−1)n 1 ∞ √ solution The series Leibniz Test, but the positive series is a divergent p-series. √ −1 3 n converges byRthe 3 = (n(n + 1)(n + 2)) . n 2. Which of the following statements is equivalentn=1 to Theorem 1? ∞ ∞ Exercise 46 in Section R is a telescoping series with the sum R = 14 . (a) According If |an | to diverges, then an also10.2, diverges. Preliminary Questions (b) If (c) If n=0 ∞ n=0 ∞ n=0 an diverges, then ∞ |an | also diverges. n=0 ∞ an converges, then n=0 |an | also converges. n=0 solution The correct answer is (b): If ∞ ∞ an diverges, then n=0 |an | also diverges. Take an = (−1)n n1 to see that n=0 statements (a) and (c) are not true in general. ∞ √ (−1)n n is an alternating series and therefore converges. Is Lathika right? 3. Lathika argues that n=1 solution No. Although ∞ √ √ (−1)n n is an alternating series, the terms an = n do not form a decreasing sequence n=1 that tends to zero. In fact, an = ∞ √ √ n is an increasing sequence that tends to ∞, so (−1)n n diverges by the Divergence n=1 Test. 4. Suppose that an is positive, decreasing, and tends to 0, and let S = ∞ (−1)n−1 an . What can we say about |S − S100 | n=1 if a101 = 10−3 ? Is S larger or smaller than S100 ? solution From the text, we know that |S − S100 | < a101 = 10−3 .Also, the Leibniz test tells us that S2N < S < S2N+1 for any N ≥ 1, so that S100 < S. Exercises 1. Show that ∞ (−1)n 2n n=0 converges absolutely. solution The positive series ∞ 1 1 is a geometric series with r = . Thus, the positive series converges, and the 2n 2 n=0 given series converges absolutely. May 23, 2011 S E C T I O N 10.4 Absolute and Conditional Convergence 671 In Exercises determine whether series converges absolutely, conditionally, or not at all. Show3–10, that the following series the converges conditionally: 3. ∞ (−1)n−1 n1/3 ∞ n=1 n=1 1 1 1 1 1 (−1)n−1 2/3 = 2/3 − 2/3 + 2/3 − 2/3 + · · · n 1 2 3 4 1 is positive, decreasing, and tends to zero; hence, the series solution The sequence an = 1/3 n by the Leibniz Test. However, the positive series n=1 ∞ n=1 ∞ (−1)n−1 converges n1/3 1 n1/3 is a divergent p-series, so the original series converges conditionally. 5. ∞ (−1)n−1n 4 ∞ (−1) n (1.1)n n=0 n3 + 1 n=1 solution The positive series ∞ 1 n is a convergent geometric series; thus, the original series converges abso1.1 n=0 lutely. 7. ∞ (−1)n πn ∞ sin( ) n ln n 4 n=2 n2 n=1 1 . Then a forms a decreasing sequence (note that n and ln n are both increasing functions of Let an = n ln n n ∞ ∞ (−1)n 1 converges by the Leibniz Test. However, the positive series n) that tends to zero; hence, the series n ln n n ln n solution n=2 n=2 diverges, so the original series converges conditionally. ∞ cos nπ n ∞ (−1) (ln n)2 1 n=2 n=1 1 + n solution Since cos nπ alternates between +1 and −1, 9. ∞ ∞ cos nπ (−1)n = (lnn)2 (lnn)2 n=2 n=2 This is an alternating series whose general term decreases to zero, so it converges. The associated positive series, ∞ n=2 1 (ln n)2 is a divergent series, so the original series converges conditionally. ∞ 1 ∞ 11. Let S =cos n(−1)n+1 3 . n n=1 2n n=1 (a) Calculate Sn for 1 ≤ n ≤ 10. (b) Use Eq. (2) to show that 0.9 ≤ S ≤ 0.902. solution (a) S1 = 1 1 7 S2 = 1 − 3 = = 0.875 8 2 1 S3 = S2 + 3 = 0.912037037 3 1 S4 = S3 − 3 = 0.896412037 4 1 S5 = S4 + 3 = 0.904412037 5 May 23, 2011 1 S6 = S5 − 3 = 0.899782407 6 1 S7 = S6 + 3 = 0.902697859 7 1 S8 = S7 − 3 = 0.900744734 8 1 S9 = S8 + 3 = 0.902116476 9 1 S10 = S9 − 3 = 0.901116476 10 672 C H A P T E R 10 INFINITE SERIES (b) By Eq. (2), |S10 − S| ≤ a11 = 1 , 113 so S10 − 1 1 ≤ S ≤ S10 + 3 , 113 11 or 0.900365161 ≤ S ≤ 0.901867791. ∞ (−1)n+1 Use Eq. (2) to approximate 13. Approximate to three decimal places. n4 n=1 ∞ (−1)n+1 ∞ n+1 1 n! (−1) n=1 , so that an = 4 . By Eq. (2), solution Let S = n4 n n=1 to four decimal places. |SN − S| ≤ aN+1 = 1 . (N + 1)4 To guarantee accuracy to three decimal places, we must choose N so that 1 < 5 × 10−4 (N + 1)4 or N > √ 4 2000 − 1 ≈ 5.7. The smallest value that satisfies the required inequality is then N = 6. Thus, 1 1 1 1 1 S ≈ S6 = 1 − 4 + 4 − 4 + 4 − 4 = 0.946767824. 2 3 4 5 6 −5 In Exercises Let15 and 16, find a value of N such that SN approximates the series with an error of at most 10 . If you have a CAS, compute this value of SN . ∞ n ∞ (−1)n−1 2 S= (−1)n+1 n +1 15. n=1 n(n + 2)(n + 3) n=1 Use a computer algebra system to calculate and plot the partial sums Sn for 1 ≤ n ≤ 100. Observe that the partial ∞ n+1limit. 1 (−1)the sums zigzag above and below solution Let S = , so that an = . By Eq. (2), n (n + 2) (n + 3) n (n + 2) (n + 3) n=1 |SN − S| ≤ aN+1 = 1 . (N + 1)(N + 3)(N + 4) We must choose N so that 1 ≤ 10−5 (N + 1)(N + 3)(N + 4) or (N + 1)(N + 3)(N + 4) ≥ 105 . For N = 43, the product on the left hand side is 95,128, while for N = 44 the product is 101,520; hence, the smallest value of N which satisfies the required inequality is N = 44. Thus, S ≈ S44 = 44 n=1 (−1)n+1 = 0.0656746. n(n + 2)(n + 3) In Exercises determine convergence or divergence by any method. ∞ 17–32, (−1)n+1 ln n ∞ n! n=1 7−n 17. n=0 solution This is a (positive) geometric series with r = May 23, 2011 1 < 1, so it converges. 7 S E C T I O N 10.4 19. Absolute and Conditional Convergence 673 ∞ ∞ 1 1 5n − 3n 7.5 n=1 n n=1 1 Use the Limit Comparison Test with n : 5 solution 1/(5n − 3n ) 5n 1 = lim = lim =1 n→∞ n→∞ 5n − 3n n→∞ 1 − (3/5)n 1/5n L = lim But ∞ 1 n=1 5n is a convergent geometric series. Since L = 1, the Limit Comparison Test tells us that the original series converges as well. 21. ∞ ∞ 1 n 3n4 + 12n 2 n=1 n −n n=2 solution 1 : 3n4 Use the Limit Comparison Test with L = lim n→∞ (1/(3n4 + 12n) 3n4 1 =1 = lim = lim 4 4 n→∞ n→∞ 1 + 4n− 3 1/3n 3n + 12n ∞ 1 1 = 13 ∞ n=1 n4 is a convergent p-series. Since L = 1, the Limit Comparison Test tells us that the original 3n4 series converges as well. But n=1 ∞ ∞ 1 (−1)n 2 n=1 n +21 n=1 n + 1 solution Apply the Limit Comparison Test and compare the series with the divergent harmonic series: 23. √1 n n2 +1 = lim = 1. 1 n→∞ n2 + 1 n L = lim n→∞ Because L > 0, we conclude that the series ∞ n=1 1 n2 + 1 diverges. ∞ n n 3 + (−2) ∞ nn (−1) n 5 n=1 2 n=0 n + 1 solution The series 25. ∞ n ∞ n 3 3 = 5n 5 n=1 n=1 is a convergent geometric series, as is the series ∞ ∞ 2 n (−1)n 2n = . − 5n 5 n=1 n=1 Hence, ∞ n ∞ n ∞ 3 + (−1)n 2n 3 2 n = + − 5n 5 5 n=1 n=1 n=1 also converges. 27. ∞ 3 ∞ n 2 n+1 (−1)(−1) n e−n /3 n=1 (2n + 1)! n=1 solution Consider the associated positive series ∞ 3 n2 e−n /3 . This series can be seen to converge by the Integral n=1 Test: ∞ 1 3 x 2 e−x /3 dx = lim R R→∞ 1 3 3 R 3 x 2 e−x /3 dx = − lim e−x /3 1 = e−1/3 + lim e−R /3 = e−1/3 . R→∞ The integral converges, so the original series converges absolutely. May 23, 2011 R→∞ 674 C H A P T E R 10 INFINITE SERIES 29. ∞ n ∞ (−1) 3 −n 1/2 n ne(ln n)/32 n=2 n=1 1 solution This is an alternating series with an = 1/2 . Because an is a decreasing sequence which converges n (ln n)2 ∞ (−1)n to zero, the series converges by the Leibniz Test. (Note that the series converges only conditionally, not n1/2 (ln n)2 n=2 1 absolutely; the associated positive series is eventually greater than 3/4 , which is a divergent p-series). n ∞ ∞ln n 31. 1 n1.05 n=1 n(ln n)1/4 n=2 solution Choose N so that for n ≥ N we have ln n ≤ n0.01 . Then ∞ ∞ ∞ ln n n0.01 1 ≤ = n1.05 n1.05 n1.04 n=N n=N n=N This is a convergent p-series, so by the Comparison Test, the original series converges as well. 33. Show that ∞ 1 1 1 1 1 1 1 S = − + − + − + ··· (ln n)2 2 2 3 3 4 4 n=2 converges by computing the partial sums. Does it converge absolutely? solution The sequence of partial sums is S1 = 1 2 1 =0 2 1 1 S3 = S2 + = 3 3 1 S4 = S3 − = 0 3 S2 = S1 − and, in general, ⎧ ⎨1, SN = N ⎩ 0, for odd N for even N Thus, lim SN = 0, and the series converges to 0. The positive series is N→∞ ∞ 1 1 1 1 1 1 1 + + + + + + ··· = 2 ; 2 2 3 3 4 4 n n=2 which diverges. Therefore, the original series converges conditionally, not absolutely. 35. Assumptions Matter Show by The Leibniz Test cannot be applied to counterexample that the Leibniz Test does not remain true if the sequence an tends to zero but is not assumed nonincreasing. Hint: Consider 1 1 1 1 1 1 − 1+ · · 1· 1 1 1− 1+ 1 − 1 + R = − +2 −3 +22 − 32 + 2· 3· · + 33 − n + · · · 2 4 3 8 4 16 n 2 Why not? Show that it converges by another method. solution Let 1 1 1 1 1 1 1 1 + ··· + − n+1 + · · · R= − + − + − 2 4 3 8 4 16 n+1 2 This is an alternating series with an = ⎧ 1 ⎪ ⎪ ⎨ k + 1 , n = 2k − 1 ⎪ ⎪ ⎩ 1 , 2k+1 n = 2k Note that an → 0 as n → ∞, but the sequence {an } is not decreasing. We will now establish that R diverges. May 23, 2011 S E C T I O N 10.4 Absolute and Conditional Convergence 675 For sake of contradiction, suppose that R converges. The geometric series ∞ 1 n=1 2n+1 converges, so the sum of R and this geometric series must also converge; however, R+ ∞ n=1 1 = 2n+1 ∞ 1 , n n=2 which diverges because the harmonic series diverges. Thus, the series R must diverge. 37. Prove that if whether an converges absolutely, an2conditionally: also converges. Then give an example where an is only Determine the following seriesthen converges 2 conditionally convergent and an diverges. 1 1 1 1 1 1 1 1 1 + − + ··· 1 − + − + − + − 3 2 5 3 7 4 9 5 11 solution Suppose the series an converges absolutely. Because |an | converges, we know that lim |an | = 0. n→∞ Therefore, there exists a positive integer N such that |an | < 1 for all n ≥ N. It then follows that for n ≥ N, 0 ≤ an2 = |an |2 = |an | · |an | < |an | · 1 = |an |. By the Comparison Test we can then conclude that an2 also converges. ∞ (−1)n Consider the series √ . This series converges by the Leibniz Test, but the corresponding positive series is a n n=1 ∞ ∞ ∞ 1 (−1)n . an2 is the divergent harmonic series is conditionally convergent. Now, √ n n n=1 n=1 n=1 Thus, an2 need not converge if an is only conditionally convergent. divergent p-series; that is, Further Insights and Challenges 39. Use Exercise 38 to show that the following series converges: Prove the following variant of the Leibniz Test: If {an } is a positive, decreasing sequence with lim an = 0, then n→∞ 2 1 1 2 1 1 the series − + + − + ··· + S= ln 2 ln 3 ln 4 ln 5 ln 6 ln 7 a1 + a2 − 2a3 + a4 + a5 − 2a6 + · · · 1 . Because an is solution The given series has the structure of the generic series from Exercise 38 with an = ln(n+1) is increasing and bounded by a + a , and continue as in the proof the Leibniz converges. Hint: Show that S 3N 1 2 a positive, decreasing sequence with lim an = 0, we can conclude from Exercise 38 that the given seriesofconverges. n→∞ Test. 41. Show that the following series diverges: Prove the conditional convergence of 1 1 2 1 1 1 2 1 + 3 + 1 + 1 − 1 + 3 ··· 1 − S =1+ + R =1+ 2 + 3 − 4 + 5 + 6 + 7 − 8 + ··· 2 3 4 5 6 7 8 Hint: Use the result of Exercise 40 to write S as the sum of a convergent series and a divergent series. solution Let R =1+ 1 1 3 1 1 1 3 + − + + + − + ··· 2 3 4 5 6 7 8 S =1+ 1 1 2 1 1 1 2 + − + + + − + ··· 2 3 4 5 6 7 8 and For sake of contradiction, suppose the series S converges. From Exercise 40, we know that the series R converges. Thus, the series S − R must converge; however, S−R = ∞ 1 1 1 11 + + + ··· = , 4 8 12 4 k k=1 which diverges because the harmonic series diverges. Thus, the series S must diverge. Prove that ∞ May 23, 2011 n=1 (−1)n+1 (ln n)a n a 676 C H A P T E R 10 INFINITE SERIES 43. We say that {bn } is a rearrangement of {an } if {bn } has the same terms as {an } but occurring in a different order. Show ∞ ∞ an converges absolutely, then T = bn also converges absolutely. that if {bn } is a rearrangement of {an } and S = n=1 n=1 (This result does not hold if S is only conditionally convergent.) Hint: Prove that the partial sums |bn | are bounded. n=1 It can be shown further that S = T . solution N ∞ Suppose the series S = an converges absolutely and denote the corresponding positive series by n=1 S+ = ∞ |an |. n=1 Further, let TN = N |bn | denote the Nth partial sum of the series n=1 ∞ |bn |. Because {bn } is a rearrangement of {an }, we n=1 know that 0 ≤ TN ≤ ∞ |an | = S + ; n=1 that is, the sequence {TN } is bounded. Moreover, TN+1 = N+1 |bn | = TN + |bN+1 | ≥ TN ; n=1 that is, {TN } is increasing. It follows that {TN } converges, so the series ∞ |bn | converges, which means the series n=1 ∞ bn n=1 converges absolutely. Assumptions Matter In 1829, Lejeune Dirichlet pointed out that the great French mathematician Augustin Louis Cauchy made a mistake in a published paper by improperly assuming the Limit Comparison Test to be valid 10.5 The Ratioseries. andHere Root for nonpositive are Tests Dirichlet’s two series: ∞ ∞ (−1)n (−1)n (−1)n , 1 + √ √ √ an+1 n an n n or lim n=1 ? 1. In the Ratio Test, is ρ equal to lim n=1 n→∞ an n→∞ an+1 Explain how they provide a counterexample to the Limit Comparison Test when the series are not assumed to be positive. In the Ratio Test ρ is the limit lim an+1 . solution n→∞ an Preliminary Questions 2. Is the Ratio Test conclusive for ∞ ∞ 1 1 ? Is it conclusive for ? n 2 n n=1 solution The general term of ∞ n=1 and n=1 1 1 is an = n ; thus, 2n 2 an+1 1 2n 1 a = 2n+1 · 1 = 2 , n a 1 ρ = lim n+1 = < 1. n→∞ an 2 Consequently, the Ratio Test guarantees that the series The general term of ∞ 1 n=1 ∞ 1 converges. 2n n=1 1 is an = ; thus, n n May 23, 2011 an+1 n n 1 a = n + 1 · 1 = n + 1, n S E C T I O N 10.5 and The Ratio and Root Tests 677 a n = 1. ρ = lim n+1 = lim n→∞ an n→∞ n + 1 The Ratio Test is therefore inconclusive for the series ∞ 1 . n n=1 3. Can the Ratio Test be used to show convergence if the series is only conditionally convergent? solution No. The Ratio Test can only establish absolute convergence and divergence, not conditional convergence. Exercises In Exercises 1–20, apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive. 1. ∞ 1 5n n=1 solution With an = 51n , an+1 5n 1 1 a = 5n+1 · 1 = 5 n Therefore, the series an+1 1 = < 1. and ρ = lim n→∞ an 5 ∞ 1 converges by the Ratio Test. 5n n=1 ∞ ∞1 3. (−1)n−1 n nn n=1 5n n=1 solution With an = n1n , n an+1 1 n 1 −n 1 nn 1 = = 1 + · = , a (n + 1)n+1 1 n+1 n+1 n+1 n n and a 1 ρ = lim n+1 = 0 · = 0 < 1. n→∞ an e Therefore, the series ∞ 1 converges by the Ratio Test. nn n=1 ∞ ∞ n 5. 3n + 2 n2 + 13 n=1 5n + 1 n=0 solution With an = n , n2 +1 an+1 n+1 n2 + 1 n+1 n2 + 1 = = · · , a (n + 1)2 + 1 n n n2 + 2n + 2 n and a ρ = lim n+1 = 1 · 1 = 1. n→∞ an ∞ n , the Ratio Test is inconclusive. n2 + 1 n=1 We can show that this series diverges by using the Limit Comparison Test and comparing with the divergent harmonic series. Therefore, for the series ∞ n ∞2 n 2 n100 n=1 n n=1 n solution With an = 2100 , 7. n 100 an+1 n100 n 2n+1 a = (n + 1)100 · 2n = 2 n + 1 n Therefore, the series ∞ 2n diverges by the Ratio Test. n100 n=1 May 23, 2011 and an+1 = 2 · 1100 = 2 > 1. ρ = lim n→∞ an 678 C H A P T E R 10 INFINITE SERIES ∞ 10n 3 ∞ n n2 n=1 2 3n2 n=1 n solution With an = 10n2 , 9. 2 2 n+1 an+1 2n 1 = 10 · = 10 · 2n+1 a 2 n (n+1) 10 2 n 2 Therefore, the series ∞ 10n n2 n=1 2 and a ρ = lim n+1 = 10 · 0 = 0 < 1. n→∞ an converges by the Ratio Test. ∞ n e n ∞ e nn n=1 n! n=1 n solution With an = ne n , n an+1 en+1 nn e e n 1 −n = · = = , 1 + a (n + 1)n+1 en n+1 n+1 n+1 n n 11. and a 1 ρ = lim n+1 = 0 · = 0 < 1. n→∞ an e Therefore, the series ∞ n e converges by the Ratio Test. nn n=1 ∞ n! 40 ∞ 13. n 6n n=0 n! n=1 solution With an = 6n!n , an+1 (n + 1)! 6n n+1 a = 6n+1 · n! = 6 n Therefore, the series an+1 = ∞ > 1. and ρ = lim n→∞ an ∞ n! diverges by the Ratio Test. 6n n=0 ∞ ∞1 15. n! n ln n n=2 n9 n=1 1 , solution With an = n ln n an+1 n ln n n ln n 1 a = (n + 1) ln(n + 1) · 1 = n + 1 ln(n + 1) , n and an+1 ln n = 1 · lim ρ = lim . n→∞ an n→∞ ln(n + 1) Now, lim ln n n→∞ ln(n + 1) = lim ln x x→∞ ln(x + 1) = lim x→∞ Thus, ρ = 1, and the Ratio Test is inconclusive for the series Using the Integral Test, we can show that the series ∞ n=2 ∞ n=2 1/(x + 1) x = lim = 1. x→∞ x + 1 1/x 1 . n ln n 1 diverges. n ln n ∞ 2 ∞ n 17. 1 (2n + 1)! n=1 (2n)! n=1 n2 , solution With an = (2n+1)! 2 2 an+1 1 = (n + 1) · (2n + 1)! = n + 1 , a (2n + 3)! n (2n + 3)(2n + 2) n2 n May 23, 2011 S E C T I O N 10.5 The Ratio and Root Tests and a ρ = lim n+1 = 12 · 0 = 0 < 1. n→∞ a n Therefore, the series ∞ n=1 19. n2 converges by the Ratio Test. (2n + 1)! ∞ ∞ 1 (n!)3 2n + 1 n=2 (3n)! n=1 1 , solution With an = n 2 +1 an+1 1 + 2−n 1 2n + 1 = a = 2n+1 + 1 · 1 2 + 2−n n and a 1 ρ = lim n+1 = < 1 n→∞ an 2 Therefore, the series ∞ n=2 1 converges by the Ratio Test. 2n + 1 ∞ ∞ nk 3−n converges for all exponents k. 21. Show that1 ln nn=1 n=2 solution With an = nk 3−n , an+1 (n + 1)k 3−(n+1) 1 1 k = , 1 + a = 3 n nk 3−n n and, for all k, a 1 1 ρ = lim n+1 = · 1 = < 1. n→∞ an 3 3 Therefore, the series ∞ nk 3−n converges for all exponents k by the Ratio Test. n=1 ∞ ∞n n 2 x 2 converges if |x| < 1 . 23. Show that n x n converges if |x| 2< 1. Show that solution n=1 n=1 With an = 2n x n , an+1 2n+1 |x|n+1 = 2|x| a = 2n |x|n n Therefore, ρ < 1 and the series ∞ a and ρ = lim n+1 = 2|x|. n→∞ an 2n x n converges by the Ratio Test provided |x| < 12 . n=1 ∞ n r n ∞ 25. Show that rconverges if |r| < 1. converges for all r. Show that n n=1 n! n=1 n solution With an = rn , an+1 |r|n+1 n n a = n + 1 · |r|n = |r| n + 1 n Therefore, by the Ratio Test, the series ∞ n r converges provided |r| < 1. n n=1 May 23, 2011 a and ρ = lim n+1 = 1 · |r| = |r|. n→∞ an 679 680 C H A P T E R 10 INFINITE SERIES ∞ 1 n n! ∞ n converges. Hint: Use lim = e. 1 + 27. Show that 2 n n→∞ n Is there any nvalue of k such that converges? k n=1 n n=1 solution With an = nn!n , n n an+1 1 −n n = (n + 1)! · n = = 1 + , a (n + 1)n+1 n! n+1 n n and a 1 ρ = lim n+1 = < 1. n→∞ an e Therefore, the series ∞ n! converges by the Ratio Test. nn n=1 In Exercises 28–33, assume that |an+1 /an | converges to ρ = 13 . What can you say about the convergence of the given series? 29. ∞ ∞3 n an nan n=1 n=1 solution Let bn = n3 an . Then b n + 1 3 an+1 1 3 1 ρ = lim n+1 = lim a = 1 · 3 = 3 < 1. n→∞ bn n→∞ n n Therefore, the series ∞ n3 an converges by the Ratio Test. n=1 31. ∞ ∞n 3 ann 2 an n=1 n=1 solution Let bn = 3n an . Then b 1 3n+1 an+1 = 3 · = 1. ρ = lim n+1 = lim n→∞ bn n→∞ 3n an 3 Therefore, the Ratio Test is inconclusive for the series ∞ 3n an . n=1 33. ∞ ∞2 an n 4 an n=1 n=1 solution Let bn = an2 . Then 2 2 b a 1 1 ρ = lim n+1 = lim n+1 = = < 1. n→∞ bn n→∞ an 3 9 Therefore, the series ∞ an2 converges by the Ratio Test. n=1 ∞ 1 ∞ −1 Assume an+1 /an converges to ρ = 4. Does 35. Is the Ratio that Testconclusive for the p-series ? n=1 an converge (assume that an = 0 for all n)? np n=1 solution With an = n1p , p an+1 n np 1 = = · a (n + 1)p 1 n+1 n Therefore, the Ratio Test is inconclusive for the p-series ∞ 1 . np n=1 May 23, 2011 and a ρ = lim n+1 = 1p = 1. n→∞ an S E C T I O N 10.5 The Ratio and Root Tests In Exercises 36–41, use the Root Test to determine convergence or divergence (or state that the test is inconclusive). 37. ∞ ∞1 1 nn n n=1 10 n=0 solution With an = n1n , √ n Therefore, the series an = n 1 1 = nn n and lim n→∞ √ n a = 0 < 1. n ∞ 1 converges by the Root Test. nn n=1 ∞ ∞ k k k 39. k 3k + 1 k=0 k + 10 k=0 k k solution With ak = 3k+1 , √ k a = k Therefore, the series ∞ k=0 k 3k + 1 k k k k = 3k + 1 3k + 1 and lim √ k k→∞ ak = 1 < 1. 3 k converges by the Root Test. −n2 ∞ ∞ 1 1 −n 1+ 1n+ n=4 n n=1 −n2 solution With ak = 1 + n1 , 2 √ 1 −n 1 −n n n a = = 1+ 1+ n n n 41. Therefore, the series ∞ 1+ n=4 and lim n→∞ √ n a = e−1 < 1. n 2 1 −n converges by the Root Test. n In Exercises 43–56,∞determine convergence or divergence using any method covered in the text so far. 2n2 2 diverges. Hint: Use 2n = (2n )n and n! ≤ nn . that ∞ Prove n! 2n + 4n n=1 43. 7n n=1 solution Because the series ∞ n ∞ n 2 2 = 7n 7 n=1 and n=1 ∞ n ∞ n 4 4 = 7n 7 n=1 n=1 are both convergent geometric series, it follows that ∞ n ∞ n ∞ n 2 4 2 + 4n = + 7n 7 7 n=1 n=1 n=1 also converges. 45. ∞ 3 n 3 ∞ n 5n n=1 n! n=1 3 solution The presence of the exponential term suggests applying the Ratio Test. With an = n5n , an+1 1 3 an+1 (n + 1)3 5n 1 1 3 = = · 1 = 1 < 1. · = and ρ = lim 1 + a n→∞ an 5 n 5 5 5n+1 n3 n Therefore, the series ∞ 3 n converges by the Ratio Test. 5n n=1 May 23, 2011 681 682 C H A P T E R 10 INFINITE SERIES ∞ ∞ 1 1 3 − n2 n=2 nn(ln n)3 n=2 solution This series is similar to a p-series; because 47. 1 1 1 ≈ √ = 3/2 3 n n n3 − n2 for large n, we will apply the Limit Comparison Test comparing with the p-series with p = 32 . Now, √ 1 n3 n3 −n2 L = lim = lim = 1. 1 n→∞ n→∞ n3 − n2 3/2 n The p-series with p = 32 converges and L exists; therefore, the series 49. ∞ 1 also converges. 3 2 n=2 n − n ∞ ∞−0.82 n n + 4n n=1 3n4 + 9 n=1 solution ∞ n−0.8 = n=1 ∞ 1 n0.8 n=1 so that this is a divergent p-series. 51. ∞ ∞−2n+1 4 (0.8)−n n−0.8 n=1 n=1 solution Observe ∞ 4−2n+1 = n=1 ∞ 4 · (4−2 )n = n=1 ∞ n 1 4 16 n=1 1 ; therefore, this series converges. is a geometric series with r = 16 ∞ ∞ 1 n−1 53. sin (−1) n2 √ n=1 n n=1 solution Here, we will apply the Limit Comparison Test, comparing with the p-series with p = 2. Now, L = lim n→∞ sin 12 n 1 2 n = lim u→0 sin u = 1, u where u = 12 . The p-series with p = 2 converges and L exists; therefore, the series n ∞ n=1 1 sin 2 also converges. n ∞ (−2)n ∞ √ 1 n n cos (−1) n solution Because 55. n=1 n=1 √ 2n 2x 2x ln 2 = lim 2x+1 x ln 2 = ∞ = 0, lim √ = lim √ = lim 1 n→∞ n x→∞ x x→∞ √ x→∞ 2 x the general term in the series ∞ n=1 ∞ (−2)n √ does not tend toward zero; therefore, the series diverges by the Divergence Test. n n n Further Insights and Challenges 57. n=1 ∞ n + 12 √ Proof of the Root Test Let S = an be a positive series, and assume that L = lim n an exists. n→∞ n=0 n (a) Show that S converges if L < 1. Hint: Choose R with L < R < 1 and show that an ≤ R for n sufficiently large. n Then compare with the geometric series R . (b) Show that S diverges if L > 1. May 23, 2011 S E C T I O N 10.5 solution Suppose lim n→∞ (a) If L < 1, let = √ n The Ratio and Root Tests 683 an = L exists. 1−L . By the definition of a limit, there is a positive integer N such that 2 √ − ≤ n an − L ≤ for n ≥ N. From this, we conclude that 0≤ √ n a ≤ L+ n for n ≥ N. Now, let R = L + . Then R =L+ L+1 1+1 1−L = < = 1, 2 2 2 and 0≤ for n ≥ N. Because 0 ≤ R < 1, the series the Comparison Test. Therefore, the series √ n ∞ an ≤ R or 0 ≤ an ≤ R n R n is a convergent geometric series, so the series n=N ∞ ∞ an converges by n=N an also converges. n=0 (b) If L > 1, let = L−1 . By the definition of a limit, there is a positive integer N such that 2 √ − ≤ n an − L ≤ for n ≥ N. From this, we conclude that L− ≤ √ n a n for n ≥ N. Now, let R = L − . Then R =L− L+1 1+1 L−1 = > = 1, 2 2 2 and R≤ for n ≥ N. Because R > 1, the series Comparison Test. Therefore, the series ∞ √ n a n or R n ≤ an R n is a divergent geometric series, so the series n=N ∞ ∞ an diverges by the n=N an also diverges. n=0 ∞ cn n! Ratio Test not apply, but verify convergence using the Comparison Test for the series , where c isdoes a constant. 59. LetShow S = that the nn n=1 1 1 1 1 1 + 3 +if |c| + + ··· (a) Prove that S converges absolutely if |c| < e + and 2diverges 4 > 2e.5 2 3 2 3 n √ e n! (b) It is known that lim n+1/2 = 2π. Verify this numerically. n→∞ n (c) Use the Limit Comparison Test to prove that S diverges for c = e. solution n (a) With an = cnnn! , n an+1 |c|n+1 (n + 1)! nn 1 −n n = · = |c| 1 + , = |c| a |c|n n! n+1 n (n + 1)n+1 n and an+1 = |c|e−1 . ρ = lim n→∞ an Thus, by the Ratio Test, the series ∞ n c n! converges when |c|e−1 < 1, or when |c| < e. The series diverges when nn n=1 |c| > e. May 23, 2011 684 C H A P T E R 10 INFINITE SERIES √ n n! (b) The table below lists the value of en+1/2 for several increasing values of n. Since 2π = 2.506628275, the numerical n evidence verifies that lim en n! n→∞ nn+1/2 = √ 2π. n 100 1000 10000 100000 en n! nn+1/2 2.508717995 2.506837169 2.506649163 2.506630363 (c) With c = e, the series S becomes ∞ n e n! . Using the result from part (b), nn n=1 en n! nn √ en n! L = lim √ = lim n+1/2 = 2π. n→∞ n→∞ n n Because the series ∞ ∞ n √ e n! n diverges by the Divergence Test and L > 0, we conclude that diverges by the Limit nn n=1 n=1 Comparison Test. 10.6 Power Series Preliminary Questions 1. Suppose that an x n converges for x = 5. Must it also converge for x = 4? What about x = −3? solution The power series an x n is centered at x = 0. Because the series converges for x = 5, the radius of convergence must be at least 5 and the series converges absolutely at least for the interval |x| < 5. Both x = 4 and x = −3 are inside this interval, so the series converges for x = 4 and for x = −3. 2. Suppose that an (x − 6)n converges for x = 10. At which of the points (a)–(d) must it also converge? (a) x = 8 (b) x = 11 (c) x = 3 (d) x = 0 solution The given power series is centered at x = 6. Because the series converges for x = 10, the radius of convergence must be at least |10 − 6| = 4 and the series converges absolutely at least for the interval |x − 6| < 4, or 2 < x < 10. (a) x = 8 is inside the interval 2 < x < 10, so the series converges for x = 8. (b) x = 11 is not inside the interval 2 < x < 10, so the series may or may not converge for x = 11. (c) x = 3 is inside the interval 2 < x < 10, so the series converges for x = 2. (d) x = 0 is not inside the interval 2 < x < 10, so the series may or may not converge for x = 0. 3. What is the radius of convergence of F (3x) if F (x) is a power series with radius of convergence R = 12? solution If the power series F (x) has radius of convergence R = 12, then the power series F (3x) has radius of convergence R = 12 3 = 4. 4. The power series F (x) = ∞ nx n has radius of convergence R = 1. What is the power series expansion of F (x) n=1 and what is its radius of convergence? solution We obtain the power series expansion for F (x) by differentiating the power series expansion for F (x) term-by-term. Thus, F (x) = ∞ n2 x n−1 . n=1 The radius of convergence for this series is R = 1, the same as the radius of convergence for the series expansion for F (x). May 23, 2011 S E C T I O N 10.6 Exercises 1. Use the Ratio Test to determine the radius of convergence R of Power Series 685 ∞ n x . Does it converge at the endpoints x = ±R? 2n n=0 solution With an = xn , 2n an+1 |x|n+1 2n |x| a = 2n+1 · |x|n = 2 n a |x| and ρ = lim n+1 = . n→∞ an 2 |x| By the Ratio Test, the series converges when ρ = |x| 2 < 1, or |x| < 2, and diverges when ρ =2 > 1, or |x| > 2. n The radius of convergence is therefore R = 2. For x = −2, the left endpoint, the series becomes ∞ n=0 (−1) , which is ∞ divergent. For x = 2, the right endpoint, the series becomes n=0 1, which is also divergent. Thus the series diverges at both endpoints. 3. Show that the power series (a)–(c) have the same radius of convergence. Then show that (a) diverges at both endpoints, ∞ n (b) converges at one endpoint but diverges atxthe other, and (c) converges at both endpoints. Use the Ratio Test to show that √ n has radius of convergence R = 2. Then determine whether it converges ∞ n ∞ ∞ n2 x xn xn n=1 (b) (c) (a) at the nendpoints R = ±2. 3 n3n n2 3n n=1 n=1 n=1 solution n (a) With an = x3n , x x x n+1 3n an+1 = lim n+1 · n = lim = ρ = lim n→∞ an n→∞ 3 x n→∞ 3 3 Then ρ < 1 if |x| < 3, so that the radius of convergence is R = 3. For the endpoint x = 3, the series becomes ∞ n ∞ 3 = 1, 3n n=1 n=1 which diverges by the Divergence Test. For the endpoint x = −3, the series becomes ∞ ∞ (−3)n = (−1)n , 3n n=1 n=1 which also diverges by the Divergence Test. n x , (b) With an = n3 n n+1 n x an+1 x n3 n x = . = lim · = lim ρ = lim n n→∞ an n→∞ (n + 1)3n+1 n→∞ x 3 n+1 3 Then ρ < 1 when |x| < 3, so that the radius of convergence is R = 3. For the endpoint x = 3, the series becomes ∞ ∞ 3n 1 , = n3n n n=1 n=1 which is the divergent harmonic series. For the endpoint x = −3, the series becomes ∞ ∞ (−3)n (−1)n = , n3n n n=1 n=1 which converges by the Leibniz Test. n (c) With an = x2 n , n 3 2 x an+1 n2 3n n x n+1 x · n = lim = lim ρ = lim = n→∞ an n→∞ (n + 1)2 3n+1 n→∞ x 3 n+1 3 Then ρ < 1 when |x| < 3, so that the radius of convergence is R = 3. For the endpoint x = 3, the series becomes ∞ ∞ 3n 1 = , n2 3n n2 n=1 May 23, 2011 n=1 686 C H A P T E R 10 INFINITE SERIES which is a convergent p-series. For the endpoint x = −3, the series becomes ∞ ∞ (−3)n (−1)n = , n2 3n n2 n=1 n=1 which converges by the Leibniz Test. ∞ Repeat for the following 5. Show that Exercise nn x n3diverges for all x =series: 0. ∞ ∞ n (xn=0 − 5) (x − 5)n (b) (a) n n n 9 an = n x , and assuming x = 0, n9n solution With n=1 (c) n=1 ∞ (x − 5)n n2 9n n=1 n (n + 1)n+1 x n+1 an+1 =∞ = lim x 1 + 1 ρ = lim = lim (n + 1) n n n→∞ an n→∞ n→∞ n x n ρ < 1 only if x = 0, so that the radius of convergence is therefore R = 0. In other words, the power series converges only for x = 0. ∞ 2n √ ∞ x 7. Use the Ratio Test to show that nn has radius of convergence R = 3. 3 converge? n!x For which values of x does n=0 x 2n n=0 solution With an = n , 3 x 2(n+1) 3n x2 x2 an+1 = lim · = lim ρ = lim = n→∞ an n→∞ 3n+1 x 2n n→∞ 3 3 Then ρ < 1 when |x 2 | < 3, or x = √ √ 3, so the radius of convergence is R = 3. In Exercises 9–34, find interval of convergence. ∞ the x 3n+1 Show that has radius of convergence R = 4. ∞ 64n n=0 nx n 9. n=0 solution With an = nx n , (n + 1)x n+1 an+1 n + 1 = |x| ρ = lim = lim = lim x n→∞ an n→∞ n→∞ nx n n Then ρ < 1 when |x| < 1, so that the radius of convergence is R = 1, and the series converges absolutely on the interval ∞ n, which diverges by the Divergence Test. |x| < 1, or −1 < x < 1. For the endpoint x = 1, the series becomes For the endpoint x = −1, the series becomes ∞ ∞ n=0 (−1)n n, which also diverges by the Divergence Test. Thus, the series n=1 nx n converges for −1 < x < 1 and diverges elsewhere. n=0 11. ∞ x 2n+1 ∞ (−1)2nn nn x2 n n=1 n n=1 x 2n+1 , 2n n x 2(n+1)+1 x2 n x 2 2n n · · ρ = lim n+1 = = lim n→∞ 2 (n + 1) x 2n+1 n→∞ 2 n + 1 2 solution With an = (−1)n √ √ Then ρ < 1 when |x| < 2, so the radius of convergence is R = 2, and the series converges absolutely on the interval √ √ ∞ ∞ √ √ √ − 2 2 = , which converges (−1)n (−1)n+1 − 2 < x < 2. For the endpoint x = − 2, the series becomes n n n=1 n=1 √ ∞ √ 2 n which also converges by the Leibniz test. (−1) by the Leibniz test. For the endpoint x = 2, the series becomes n Thus the series ∞ n=1 ∞ n=0 n=1 √ √ x 2n+1 (−1)n n converges for − 2 ≤ x ≤ 2 and diverges elsewhere. 2 n n (−1)n n x 2n 4 May 23, 2011 S E C T I O N 10.6 13. Power Series 687 ∞ n x n5 n=4 n solution With an = x 5 , n 5 x n+1 5 an+1 n n = lim ρ = lim · = lim x = |x| n→∞ an n→∞ (n + 1)5 x n n→∞ n+1 Then ρ < 1 when |x| < 1, so the radius of convergence is R = 1, and the series converges absolutely on the interval ∞ 1 , which is a convergent p-series. For the |x| < 1, or −1 < x < 1. For the endpoint x = 1, the series becomes n5 n=1 ∞ ∞ (−1)n , which converges by the Leibniz Test. Thus, the series endpoint x = −1, the series becomes n5 n=1 n=4 for −1 ≤ x ≤ 1 and diverges elsewhere. 15. xn converges n5 ∞ n ∞x (n!)n27 x n n=0 n=8 n solution With an = x 2 , (n!) 2 x n+1 an+1 1 (n!)2 · n = lim x = lim ρ = lim =0 n→∞ an n→∞ ((n + 1)!)2 x n→∞ n+1 ρ < 1 for all x, so the radius of convergence is R = ∞, and the series converges absolutely for all x. 17. ∞ (2n)!n n ∞ 8x (n!)3 x n n=0 n! n=0 n solution With an = (2n)!x3 , and assuming x = 0, (n!) (2(n + 1))!x n+1 a (2n + 2)(2n + 1) (n!)3 x · = lim ρ = lim n+1 = lim n→∞ an n→∞ ((n + 1)!)3 (2n)!x n n→∞ (n + 1)3 4n−1 + 6n−1 + 2n−3 4n2 + 6n + 2 = lim x 3 =0 = lim x n→∞ n + 3n2 + 3n + 1 n→∞ 1 + 3n−1 + 3n−2 + n−3 Then ρ < 1 for all x, so the radius of convergence is R = ∞, and the series converges absolutely for all x. 19. ∞ (−1)n x nn ∞ 4 2+1 x 2n−1 n=0 n(2n + 1)! n=0 n n √ x , solution With an = (−1) 2 n +1 (−1)n+1 x n+1 2 + 1 an+1 n = lim ρ = lim · n→∞ an n→∞ n2 + 2n + 2 (−1)n x n 2+1 2 n 1 + 1/n n2 + 1 = lim x = lim x = lim x 2 2 n→∞ n + 2n + 2 n→∞ 1 + 2/n + 2/n n2 + 2n + 2 n→∞ = |x| Then ρ < 1 when |x| < 1, so the radius of convergence is R = 1, and the series converges absolutely on the interval ∞ (−1)n −1 < x < 1. For the endpoint x = 1, the series becomes , which converges by the Leibniz Test. For the 2 n=1 n + 1 ∞ 1 , which diverges by the Limit Comparison Test comparing with the endpoint x = −1, the series becomes 2 n=1 n + 1 ∞ (−1)n x n converges for −1 < x ≤ 1 and diverges elsewhere. divergent harmonic series. Thus, the series 2 n=0 n + 1 ∞ n=0 xn n4 + 2 May 23, 2011 688 C H A P T E R 10 INFINITE SERIES 21. ∞ x 2n+1 3n + 1 n=15 solution With an = x 2n+1 , 3n + 1 x 2n+3 3n + 1 an+1 2 3n + 1 = |x 2 | ρ = lim · = lim = lim x n→∞ an n→∞ 3n + 4 x 2n+1 n→∞ 3n + 4 Then ρ < 1 when |x 2 | < 1, so the radius of convergence is R = 1, and the series converges absolutely for −1 < x < 1. ∞ 1 , which diverges by the Limit Comparison Test comparing For the endpoint x = 1, the series becomes 3n + 1 n=15 with the divergent harmonic series. For the endpoint x = −1, the series becomes ∞ n=15 −1 , which also diverges by 3n + 1 the Limit Comparison Test comparing with the divergent harmonic series. Thus, the series ∞ x 2n+1 converges for 3n + 1 n=15 −1 < x < 1 and diverges elsewhere. ∞ xn ∞ xn ln n n=2 n − 4 ln n n=1 n solution With an = lnx n , 23. x n+1 an+1 ln(n + 1) 1/(n + 1) ln n n = lim ρ = lim · = lim = lim = |x| x x x = lim n→∞ an n→∞ ln(n + 1) x n n→∞ ln n n→∞ 1/n n→∞ n + 1 using L’Hôpital’s rule. Then ρ < 1 when |x| < 1, so the radius of convergence is 1, and the series converges absolutely ∞ 1 on the interval |x| < 1, or −1 < x < 1. For the endpoint x = 1, the series becomes . Because ln1n > n1 and ln n n=2 ∞ 1 is the divergent harmonic series, the endpoint series diverges by the Comparison Test. For the endpoint x = −1, n n=2 the series becomes ∞ ∞ (−1)n xn , which converges by the Leibniz Test. Thus, the series converges for −1 ≤ x < 1 ln n ln n n=2 n=2 and diverges elsewhere. ∞ ∞ 3n+2n n(x x− 3) n=1 ln n n=2 solution With an = n(x − 3)n , 25. (n + 1)(x − 3)n+1 an+1 = lim (x − 3) · n + 1 = |x − 3| ρ = lim = lim n n→∞ an n→∞ n→∞ n(x − 3) n Then ρ < 1 when |x − 3| < 1, so the radius of convergence is 1, and the series converges absolutely on the interval ∞ |x − 3| < 1, or 2 < x < 4. For the endpoint x = 4, the series becomes n, which diverges by the Divergence Test. For the endpoint x = 2, the series becomes ∞ ∞ n=1 (−1)n n, which also diverges by the Divergence Test. Thus, the series n=1 n(x − 3)n converges for 2 < x < 4 and diverges elsewhere. n=1 ∞ ∞ n 5n nn (−1)(−5) n (x(x−−7)3) n=1 n2 n=1 solution With an = (−1)n n5 (x − 7)n , (−1)n+1 (n + 1)5 (x − 7)n+1 an+1 (n + 1)5 ρ = lim = lim = lim (x − 7) · n→∞ an n→∞ n→∞ (−1)n n5 (x − 7)n n5 n5 + . . . = lim (x − 7) · = |x − 7| n→∞ n5 27. May 23, 2011 Power Series S E C T I O N 10.6 689 Then ρ < 1 when |x − 7| < 1, so the radius of convergence is 1, and the series converges absolutely on the interval ∞ ∞ (−1)2n n5 = n5 , which diverges by the |x − 7| < 1, or 6 < x < 8. For the endpoint x = 6, the series becomes Divergence Test. For the endpoint x = 8, the series becomes Thus, the series ∞ ∞ n=1 n=1 (−1)n n5 , which also diverges by the Divergence Test. n=1 (−1)n n5 (x − 7)n converges for 6 < x < 8 and diverges elsewhere. n=1 29. ∞ n 2 ∞ (x + 3)n 3n 27n (x − 1)3n+2 n=1 n=0 n n solution With an = 2 (x+3) , 3n 2n+1 (x + 3)n+1 an+1 3n 2(x + 3) · 3n · = lim = lim ρ = lim n→∞ an n→∞ 3(n + 1) 2n (x + 3)n n→∞ 3n + 3 1 = |2(x + 3)| = lim 2(x + 3) · n→∞ 1 + 1/n Then ρ < 1 when |2(x + 3)| < 1, so when |x + 3| < 12 . Thus the radius of convergence is 12 , and the series converges ∞ 1 absolutely on the interval |x + 3| < 12 , or − 72 < x < − 52 . For the endpoint x = − 52 , the series becomes , 3n n=1 which diverges because it is a multiple of the divergent harmonic series. For the endpoint x = − 72 , the series becomes ∞ ∞ n (−1)n 2 , which converges by the Leibniz Test. Thus, the series (x + 3)n converges for − 72 ≤ x < − 52 and 3n 3n n=1 n=1 diverges elsewhere. 31. ∞ (−5)n ∞ (x4) +n10)n (x − n! n=0 n! n=0 n n solution With an = (−5) n! (x + 10) , (−5)n+1 (x + 10)n+1 an+1 n! 5(x + 10) 1 = 0 ρ = lim · = lim = lim n n n→∞ an n→∞ n→∞ (n + 1)! (−5) (x + 10) n Thus ρ < 1 for all x, so the radius of convergence is infinite, and ∞ (−5)n (x + 10)n converges for all x. n! n=0 33. ∞ ∞n e (x − 2)n n n! (x + 5) n=12 n=10 solution With an = en (x − 2)n , en+1 (x − 2)n+1 an+1 ρ = lim = lim = lim |e(x − 2)| = |e(x − 2)| n→∞ a n→∞ en (x − 2)n n→∞ n Thus ρ < 1 when |e(x − 2)| < 1, so when |x − 2| < e−1 . Thus the radius of convergence is e−1 , and the series converges absolutely on the interval |x − 2| < e−1 , or 2 − e−1 < x < 2 + e−1 . For the endpoint x = 2 + e−1 , the series becomes ∞ ∞ 1, which diverges by the Divergence Test. For the endpoint x = 2 − e−1 , the series becomes (−1)n , which also n=1 diverges by the Divergence Test. Thus, the series ∞ n=12 elsewhere. ∞ (x + 4)n (n ln n)2 n=2 May 23, 2011 n=1 en (x − 2)n converges for 2 − e−1 < x < 2 + e−1 and diverges 690 C H A P T E R 10 INFINITE SERIES In Exercises 35–40, use Eq. (2) to expand the function in a power series with center c = 0 and determine the interval of convergence. 1 1 − 3x solution Substituting 3x for x in Eq. (2), we obtain 35. f (x) = ∞ ∞ 1 = (3x)n = 3n x n . 1 − 3x n=0 n=0 This series is valid for |3x| < 1, or |x| < 13 . 1 37. f (x) = 1 f (x) 3=− x 1 + 3x solution First write 1 1 1 = · . 3−x 3 1 − x3 Substituting x3 for x in Eq. (2), we obtain ∞ ∞ x n xn 1 = = ; x 1− 3 3 3n n=0 n=0 Thus, ∞ ∞ xn 1 xn 1 = = . n n+1 3−x 3 3 3 n=0 n=0 This series is valid for |x/3| < 1, or |x| < 3. 1 39. f (x) = 1 f (x) 1=+ x 2 4 + 3x solution Substituting −x 2 for x in Eq. (2), we obtain ∞ ∞ 1 2 )n = = (−x (−1)n x 2n 1 + x2 n=0 n=0 This series is valid for |x| < 1. 41. Use the equalities 1 f (x) = 16 + 2x 3 − 13 1 1 = = 1−x −3 − (x − 4) 1 + x−4 3 to show that for |x − 4| < 3, ∞ (x − 4)n 1 = (−1)n+1 n+1 1−x 3 n=0 solution Substituting − x−4 3 for x in Eq. (2), we obtain 1 1 + x−4 3 = ∞ ∞ x−4 n (x − 4)n = (−1)n . − 3 3n n=0 n=0 Thus, ∞ ∞ n (x − 4)n 1 1 n+1 (x − 4) . =− (−1)n = (−1) 1−x 3 3n 3n+1 n=0 n=0 This series is valid for | − x−4 3 | < 1, or |x − 4| < 3. 43. Use the method of Exercise 41 to expand 1/(4 − x) in a power series with center c = 5. Determine the interval of Use the method of Exercise 41 to expand 1/(1 − x) in power series with centers c = 2 and c = −2. Determine convergence. the interval of convergence. solution First write 1 1 1 =− . = −1 − (x − 5) 1 + (x − 5) 4−x May 23, 2011 S E C T I O N 10.6 Power Series 691 Substituting −(x − 5) for x in Eq. (2), we obtain ∞ ∞ 1 = (−(x − 5))n = (−1)n (x − 5)n . 1 + (x − 5) n=0 n=0 Thus, ∞ ∞ 1 =− (−1)n (x − 5)n = (−1)n+1 (x − 5)n . 4−x n=0 n=0 This series is valid for | − (x − 5)| < 1, or |x − 5| < 1. 45. Apply integration to the expansion Find a power series that converges only for x in [2, 6). ∞ 1 = (−1)n x n = 1 − x + x 2 − x 3 + · · · 1+x n=0 to prove that for −1 < x < 1, ln(1 + x) = ∞ x2 x3 x4 (−1)n−1 x n =x− + − + ··· n 2 3 4 n=1 solution To obtain the first expansion, substitute −x for x in Eq. (2): ∞ ∞ 1 = (−x)n = (−1)n x n . 1+x n=0 n=0 This expansion is valid for | − x| < 1, or −1 < x < 1. Upon integrating both sides of the above equation, we find ln(1 + x) = dx = 1+x ⎛ ⎝ ∞ ⎞ (−1)n x n ⎠ dx. n=0 Integrating the series term-by-term then yields ln(1 + x) = C + ∞ n=0 (−1)n x n+1 . n+1 To determine the constant C, set x = 0. Then 0 = ln(1 + 0) = C. Finally, ln(1 + x) = ∞ n=0 (−1)n ∞ x n+1 xn = (−1)n−1 . n+1 n n=1 47. Let F (x) = (x + 1) ln(1 + x) − x. Use the result of Exercise 45 to prove that (a) Apply integration to the result of Exercise 45 to prove that for −1 < x < 1, 1 1 1 1 3 ∞ + − + ··· ln = − n+1 2 n+1 2F (x)2 = 2 · 2(−1) 3 · 23x 4 · 24 n(n + 1) Use your knowledge of alternating series to findn=1 an N such that the partial sum SN approximates ln 32 to within an 1 −3 . Confirm using a calculator to compute both SN and ln 32 . error of atatmost to prove (b) Evaluate x = 10 2 1 1 1 1 3 3 1 − + − + ··· ln − = 2 2 2 4 · 5 · 25 1 · 2 · 22 2 · 3 · 23 3 · 4 · 24 (c) Use a calculator to verify that the partial sum S4 approximates the left-hand side with an error no greater than the term a5 of the series. solution (a) Note that ln(x + 1) dx = (x + 1) ln(x + 1) − x + C May 23, 2011 692 C H A P T E R 10 INFINITE SERIES Then integrating both sides of the result of Exercise 45 gives (x + 1) ln(x + 1) − x = ln(x + 1) dx = ∞ (−1)n−1 x n dx n n=1 For −1 < x < 1, which is the interval of convergence of the series in Exercise 45, therefore, we can integrate term by term to get (x + 1) ln(x + 1) − x = ∞ ∞ ∞ (−1)n−1 (−1)n−1 x n+1 x n+1 x n dx = (−1)n+1 · +C = +C n n n+1 n(n + 1) n=1 n=1 n=1 (noting that (−1)n−1 = (−1)n+1 ). To determine C, evaluate both sides at x = 0 to get 0 = ln 1 − 0 = 0 + C so that C = 0 and we get finally (x + 1) ln(x + 1) − x = ∞ (−1)n+1 n=1 x n+1 n(n + 1) (b) Evaluating the result of part(a) at x = 12 gives ∞ 1 3 3 1 ln − = (−1)n+1 2 2 2 n(n + 1)2n+1 n=1 1 1 1 1 = − + − + ... 2 3 4 4 · 5 · 25 1·2·2 2·3·2 3·4·2 (c) S4 = 1 1 1 1 − + − = 0.1078125 4 · 5 · 25 1 · 2 · 22 2 · 3 · 23 3 · 4 · 24 1 ≈ 0.0005208 a5 = 5 · 6 · 26 3 3 1 ln − ≈ 0.10819766 2 2 2 and S4 − 3 ln 3 − 1 ≈ 0.0003852 < a5 2 2 2 49. Use the result of Example 7 to show that Prove that for |x| < 1, x2 x4 x6 x8 dx− F (x) = + x 5 −x 9 + · · · + = x − 1 · 42 3 · 4 5 · 6 7 ·− 8 ··· 5 9 x +1 is an antiderivative of f (x) = tan−1 x satisfying 1/2 F (0) = 0. What is the radius of convergence of this power series? Use the first two terms to approximate 0 dx/(x 4 + 1) numerically. Use the fact that you have an alternating series solution For the −1 error <x< which is the interval convergence to show that in 1, this approximation is atofmost 0.00022. for the power series for arctangent, we can integrate term-by-term, so integrate that power series to get F (x) = tan−1 x dx = ∞ ∞ x 2n+2 (−1)n x 2n+1 dx = (−1)n 2n + 1 (2n + 1)(2n + 2) n=0 = x2 1·2 − x4 3·4 + x6 5·6 n=0 − x8 7·8 + ··· + C If we assume F (0) = 0, then we have C = 0. The radius of convergence of this power series is the same as that of the original power series, which is 1. ∞ n 51. Evaluate n . Hint: Use differentiation to show that Verifyn=1 that2 function F (x) = x tan−1 x − 12 log(x 2 + 1) is an antiderivative of f (x) = tan−1 x satisfying ∞ F (0) = 0. Then use the result of Exercise 49 with x = π6n−1 to show that nx (for |x| < 1) (1 − x)−2 = 1 n=1 1 π 1 1 1 4 − + − + ··· √ − ln = 1 · 2(3) 3 · 4(32 ) 5 · 6(33 ) 7 · 8(34 ) 6 3 2 3 Use a calculator to compare the value of the left-hand side with the partial sum S4 of the series on the right. May 23, 2011 S E C T I O N 10.6 solution Power Series 693 Differentiate both sides of Eq. (2) to obtain ∞ 1 = nx n−1 . (1 − x)2 n=1 Setting x = 12 then yields ∞ n=1 n = 2n−1 1 2 = 4. 1 − 12 Divide this equation by 2 to obtain ∞ n = 2. 2n n=1 53. Show that the following series converges absolutely for |x| < 1 and compute its sum: Use the power series for (1 + x 2 )−1 and differentiation to prove that for |x| < 1, F (x) = 1 − x − x 2 + x 3 − x 4 − x 5 + x 6 − x 7 − x 8 + · · · ∞ 2x Hint: Write F (x) as a sum of three geometric series with common ratio(2n)x x 3 . 2n−1 = (−1)n−1 (x 2 + 1)2 solution Because the coefficients in the power series n=1 are all ±1, we find a r = lim n+1 = 1. n→∞ an The radius of convergence is therefore R = r −1 = 1, and the series converges absolutely for |x| < 1. By Exercise 43 of Section 10.4, any rearrangement of the terms of an absolutely convergent series yields another absolutely convergent series with the same sum as the original series. Following the hint, we now rearrange the terms of F (x) as the sum of three geometric series: F (x) = 1 + x 3 + x 6 + · · · − x + x 4 + x 7 + · · · − x 2 + x 5 + x 8 + · · · = ∞ (x 3 )n − n=0 ∞ n=0 x(x 3 )n − ∞ x 2 (x 3 )n = n=0 1 x x2 1 − x − x2 − − = . 1 − x3 1 − x3 1 − x3 1 − x3 ∞ n2 x converges. n! n=1 1 + 2x = 1 + x − 2x 2 + x 3 + x 4 − 2x 5 + x 6 + x 7 − 2x 8 + · · · n2 x 1 + solution With an = n! , x + x 2 Show that for < 1,that 55. Find all values of |x| x such Hint: Use the hint from Exercise 53. 2 an+1 |x|(n+1) n! |x|2n+1 = · . = a 2 (n + 1)! |x|n n+1 n if |x| ≤ 1, then |x|2n+1 = 0, n→∞ n + 1 lim and the series converges absolutely. On the other hand, if |x| > 1, then |x|2n+1 = ∞, n→∞ n + 1 lim and the series diverges. Thus, ∞ n2 x converges for −1 ≤ x ≤ 1 and diverges elsewhere. n! n=1 ∞ n satisfying Find all values such series converges: equation y = −y with initial condition y(0) = 1. an xfollowing the differential 57. Find a power seriesofPx(x) = that the n=0 F (x) = 1 + 3x + x 2 + 27x 3 + x 4 + 243x 5 + · · · Then use Theorem 1 of Section 5.8 to conclude that P (x) = e−x . ∞ solution Let P (x) = an x n and note that P (0) = a0 ; thus, to satisfy the initial condition P (0) = 1, we must take a0 = 1. Now, n=0 P (x) = ∞ n=1 May 23, 2011 nan x n−1 , 694 C H A P T E R 10 INFINITE SERIES so P (x) + P (x) = ∞ ∞ nan x n−1 + n=1 an x n = n=0 ∞ ! (n + 1)an+1 + an x n . n=0 In order for this series to be equal to zero, the coefficient of x n must be equal to zero for each n; thus (n + 1)an+1 + an = 0 or an+1 = − an . n+1 Starting from a0 = 1, we then calculate a a1 = − 0 = −1; 1 1 a a2 = − 1 = ; 2 2 a 1 1 a3 = − 2 = − = − ; 3 6 3! and, in general, an = (−1)n 1 . n! Hence, ∞ P (x) = (−1)n n=0 xn . n! The solution to the initial value problem y = −y, y(0) = 1 is y = e−x . Because this solution is unique, it follows that P (x) = ∞ (−1)n n=0 xn = e−x . n! 59. Use the power series for y = ex to show that x4 x6 x2 + − + · · · .1 Let C(x) = 1 − 1 1 1 2! 4! 6! = − + − ··· 2! 3! 4! (a) Show that C(x) has an infinite radius ofe convergence. −1 to=within that C(x) and f (x) series = costo x are y partial = −ysum withSinitial conditions y(0) 1, y (0) = 0. Use (b) yourProve knowledge of alternating findboth an Nsolutions such thatofthe an error N approximates e −3 −1 This initial value problem has a unique solution, so we have C(x) = cos x for all x. of at most 10 . Confirm this using a calculator to compute both SN and e . solution Recall that the series for ex is ∞ n x2 x3 x4 x =1+x+ + + + ··· . n! 2! 3! 4! n=0 Setting x = −1 yields e−1 = 1 − 1 + 1 1 1 1 1 1 − + − +··· = − + − +··· . 2! 3! 4! 2! 3! 4! 1 . The error in approximating e−1 with the partial sum S is therefore This is an alternating series with an = (n+1)! N bounded by |SN − e−1 | ≤ aN+1 = 1 . (N + 2)! To make the error at most 10−3 , we must choose N such that 1 ≤ 10−3 (N + 2)! or (N + 2)! ≥ 1000. For N = 4, (N + 2)! = 6! = 720 < 1000, but for N = 5, (N + 2)! = 7! = 5040; hence, N = 5 is the smallest value that satisfies the error bound. The corresponding approximation is S5 = 1 1 1 1 1 − + − + = 0.368055555 2! 3! 4! 5! 6! Now, e−1 = 0.367879441, so |S5 − e−1 | = 1.761 × 10−4 < 10−3 . Let P (x) = an x n be a power series solution to y = 2xy with initial condition y(0) = 1. n=0 (a) ShowMay that23, the2011 odd coefficients a2k+1 are all zero. (b) Prove that a2k = a2k−2 /k and use this result to determine the coefficients a2k . S E C T I O N 10.6 Power Series 695 61. Find a power series P (x) satisfying the differential equation y − xy + y = 0 9 with initial condition y(0) = 1, y (0) = 0. What is the radius of convergence of the power series? ∞ solution Let P (x) = an x n . Then n=0 P (x) = ∞ nan x n−1 and P (x) = n=1 ∞ n(n − 1)an x n−2 . n=2 Note that P (0) = a0 and P (0) = a1 ; in order to satisfy the initial conditions P (0) = 1, P (0) = 0, we must have a0 = 1 and a1 = 0. Now, P (x) − xP (x) + P (x) = ∞ n(n − 1)an x n−2 − n=2 = ∞ ∞ nan x n + n=1 (n + 2)(n + 1)an+2 x n − n=0 ∞ n=0 ∞ nan x n + n=1 = 2a2 + a0 + ∞ an x n ∞ an x n n=0 ! (n + 2)(n + 1)an+2 − nan + an x n . n=1 In order for this series to be equal to zero, the coefficient of x n must be equal to zero for each n; thus, 2a2 + a0 = 0 and (n + 2)(n + 1)an+2 − (n − 1)an = 0, or 1 a2 = − a 0 2 and an+2 = n−1 an . (n + 2)(n + 1) Starting from a1 = 0, we calculate a3 = 1−1 a1 = 0; (3)(2) a5 = 2 a3 = 0; (5)(4) a7 = 4 a5 = 0; (7)(6) and, in general, all of the odd coefficients are zero. As for the even coefficients, we have a0 = 1, a2 = − 12 , a4 = 1 1 a2 = − ; (4)(3) 4! a6 = 3 3 a4 = − ; (6)(5) 6! a8 = 5 15 a6 = − (8)(7) 8! and so on. Thus, 1 3 15 1 P (x) = 1 − x 2 − x 4 − x 6 − x 8 − · · · 2 4! 6! 8! To determine the radius of convergence, treat this as a series in the variable x 2 , and observe that a2k+2 2k − 1 = lim = 0. r = lim k→∞ a2k k→∞ (2k + 2)(2k + 1) Thus, the radius of convergence is R = r −1 = ∞. 63. Prove that Find a power series satisfying Eq. (9) with initial condition y(0) = 0, y (0) = 1. ∞ (−1)k x 2k+2 J2 (x) = 2k+2 2 k! (k + 3)! k=0 is a solution of the Bessel differential equation of order 2: x 2 y + xy + (x 2 − 4)y = 0 May 23, 2011 696 C H A P T E R 10 INFINITE SERIES solution Let J2 (x) = ∞ k=0 (−1)k x 2k+2 . Then 22k+2 k! (k + 2)! ∞ (−1)k (k + 1) k=0 22k+1 k! (k + 2)! J2 (x) = x 2k+1 ∞ (−1)k (k + 1)(2k + 1) 2k x 22k+1 k! (k + 2)! J2 (x) = k=0 and x 2 J2 (x) + xJ2 (x) + (x 2 − 4)J2 (x) = ∞ ∞ (−1)k (k + 1)(2k + 1) 2k+2 (−1)k (k + 1) 2k+2 + x x 22k+1 k! (k + 2)! 22k+1 k! (k + 2)! k=0 − = (−1)k k=0 22k+2 k! (k + 2)! ∞ k=0 = ∞ k=1 is it impossible to expand f (x) Further Why Insights and Challenges Explain using Theorem 2. 65. Suppose that the coefficients of F (x) = k=0 ∞ x 2k+4 − (−1)k k(k + 2) 2k+2 + x 22k k!(k + 2)! ∞ (−1)k k=0 22k k! (k + 2)! x 2k+2 ∞ (−1)k−1 k=1 22k (k − 1)! (k + 1)! x 2k+2 ∞ (−1)k (−1)k x 2k+2 − x 2k+2 = 0. 2k 2k 2 (k − 1)!(k + 1)! 2 (k − 1)!(k + 1)! k=1 = |x| as a power series that converges in an interval around x = 0? ∞ an x n are periodic; that is, for some whole number M > 0, we have n=0 aM+n = an . Prove that F (x) converges absolutely for |x| < 1 and that a + a1 x + · · · + aM−1 x M−1 F (x) = 0 1 − xM Hint: Use the hint for Exercise 53. solution Suppose the coefficients of F (x) are periodic, with aM+n = an for some whole number M and all n. The F (x) can be written as the sum of M geometric series: F (x) = a0 1 + x M + x 2M + · · · + a1 x + x M+1 + x 2M+1 + · · · + = a2 x 2 + x M+2 + x 2M+2 + · · · + · · · + aM−1 x M−1 + x 2M−1 + x 3M−1 + · · · = a0 a1 x a2 x 2 aM−1 x M−1 a0 + a1 x + a2 x 2 + · · · + aM−1 x M−1 + + + · · · + = . 1 − xM 1 − xM 1 − xM 1 − xM 1 − xM As each geometric series converges absolutely for |x| < 1, it follows that F (x) also converges absolutely for |x| < 1. ∞ Continuity of Power Series Let F (x) = an x n be a power series with radius of convergence R > 0. 10.7 Taylor Series n=0 (a) Prove the inequality Preliminary Questions |x n − y n | ≤ n|x − y|(|x|n−1 + |y|n−1 ) 1. Determine f (0) and f (0) for a function f (x) with Maclaurin series Hint: x n − y n = (x − y)(x n−1 + x n−2 y + · · · + y n−1 ). ·∞ T (x) = 3 + 2x + 12x 2 + 5x 3 + · · 2n|an |R1n converges. Hint: Show that (b) Choose R1 with 0 < R1 < R. Show that the infinite series M = n=0 solution The Maclaurin series for a function f has the form n|an |R1n < |an |x n for all n sufficiently large if R1 < x < R. (0) (c) Use Eq. (10) to show that if |x| < Rf1 and |y| <f R(0) |Ff(x) (0) − F (y)| ≤ M|x − y|. 1 , then x+ x2 + x3 + · · · f (0) + 1! 2! x. Hint: Choose 3! (d) Prove that if |x| < R, then F (x) is continuous at R1 such that |x| < R1 < R. Show that if > 0 is given, then |F (x) − F (y)| ≤ for all y such that |x − y| < δ, where δ is any positive number that is less f (0) = 5. From this latter equation, it Matching this general expression with the than /M and R1 − |x| (see Figure 6).given series, we find f (0) = 3 and 3! follows that f (0) = 30. May 23, 2011 S E C T I O N 10.7 Taylor Series 697 2. Determine f (−2) and f (4) (−2) for a function with Taylor series T (x) = 3(x + 2) + (x + 2)2 − 4(x + 2)3 + 2(x + 2)4 + · · · solution The Taylor series for a function f centered at x = −2 has the form f (−2) + f (−2) f (−2) f (−2) f (4) (−2) (x + 2) + (x + 2)2 + (x + 2)3 + (x + 2)4 + · · · 1! 2! 3! 4! Matching this general expression with the given series, we find f (−2) = 0 and it follows that f (4) (−2) = 48. f (4) (−2) = 2. From this latter equation, 4! 3. What is the easiest way to find the Maclaurin series for the function f (x) = sin(x 2 )? solution The easiest way to find the Maclaurin series for sin x 2 is to substitute x 2 for x in the Maclaurin series for sin x. 4. Find the Taylor series for f (x) centered at c = 3 if f (3) = 4 and f (x) has a Taylor expansion f (x) = ∞ (x − 3)n n n=1 Integrating the series for f (x) term-by-term gives solution ∞ (x − 3)n+1 . n(n + 1) f (x) = C + n=1 Substituting x = 3 then yields f (3) = C = 4; so f (x) = 4 + ∞ (x − 3)n+1 . n(n + 1) n=1 5. (a) (b) (c) Let T (x) be the Maclaurin series of f (x). Which of the following guarantees that f (2) = T (2)? T (x) converges for x = 2. The remainder Rk (2) approaches a limit as k → ∞. The remainder Rk (2) approaches zero as k → ∞. solution The correct response is (c): f (2) = T (2) if and only if the remainder Rk (2) approaches zero as k → ∞. Exercises 1. Write out the first four terms of the Maclaurin series of f (x) if f (0) = 3, f (0) = 2, f (0) = 4, f (0) = 12 solution The first four terms of the Maclaurin series of f (x) are f (0) + f (0)x + f (0) 2 f (0) 3 4 12 x + x = 2 + 3x + x 2 + x 3 = 2 + 3x + 2x 2 + 2x 3 . 2! 3! 2 6 In Exercises find thefour Maclaurin and find theof interval on whichatthe expansion is valid. Write3–18, out the first terms ofseries the Taylor series f (x) centered c= 3 if 3. f (x) = solution 1 1 − 2x f (3) = 1, f (3) = 2, f (3) = 12, 1 gives Substituting 2x for x in the Maclaurin series for 1−x ∞ ∞ 1 (2x)n = 2n x n . = 1 − 2x n=0 This series is valid for |2x| < 1, or |x| < 12 . May 23, 2011 n=0 f (3) = 3 698 C H A P T E R 10 INFINITE SERIES 5. f (x) = cos 3x x f (x) = solution Substituting 1 − x 4 3x for x in the Maclaurin series for cos x gives cos 3x = ∞ (−1)n n=0 ∞ (3x)2n 9n x 2n (−1)n = . (2n)! (2n)! n=0 This series is valid for all x. 7. f (x) = sin(x 2 ) f (x) = sin(2x) solution Substituting x 2 for x in the Maclaurin series for sin x gives sin x 2 = ∞ (−1)n n=0 ∞ (x 2 )2n+1 x 4n+2 = . (−1)n (2n + 1)! (2n + 1)! n=0 This series is valid for all x. 9. f (x) = ln(1 4x − x2) f (x) = e solution Substituting −x 2 for x in the Maclaurin series for ln(1 + x) gives ln(1 − x 2 ) = ∞ ∞ ∞ 2n (−1)n−1 (−x 2 )n (−1)2n−1 x 2n x = =− . n n n n=1 n=1 n=1 This series is valid for |x| < 1. 11. f (x) = tan−1 (x 2 ) −1/2 f (x) = (1 − x) solution Substituting x 2 for x in the Maclaurin series for tan−1 x gives tan−1 (x 2 ) = ∞ (−1)n n=0 ∞ (x 2 )2n+1 x 4n+2 = . (−1)n 2n + 1 2n + 1 n=0 This series is valid for |x| ≤ 1. 13. f (x) = ex−2 2 f (x) = x 2 ex solution ex−2 = e−2 ex ; thus, ex−2 = e−2 ∞ ∞ n x xn = . n! e2 n! n=0 n=0 This series is valid for all x. 15. f (x) = ln(1 − 5x) 1 − cos x f (x) Substituting = solution −5x for x in the Maclaurin series for ln(1 + x) gives x ln(1 − 5x) = ∞ ∞ ∞ n n (−1)n−1 (−5x)n (−1)2n−1 5n x n 5 x = =− . n n n n=1 n=1 n=1 This series is valid for |5x| < 1, or |x| < 15 , and for x = − 15 . 17. f (x) = sinh x f (x) = (x 2 + 2x)ex solution Recall that sinh x = 1 x (e − e−x ). 2 Therefore, ⎛ ⎞ ∞ ∞ ∞ 1 ⎝ x n (−x)n ⎠ x n − = 1 − (−1)n . sinh x = 2 n! n! 2(n!) n=0 n=0 n=0 Now, 1 − (−1)n = May 23, 2011 0, n even 2, n odd S E C T I O N 10.7 Taylor Series 699 so sinh x = ∞ 2 k=0 ∞ x 2k+1 x 2k+1 = . 2(2k + 1)! (2k + 1)! k=0 This series is valid for all x. In Exercises f (x)19–28, = coshfind x the terms through degree four of the Maclaurin series of f (x). Use multiplication and substitution as necessary. 19. f (x) = ex sin x solution Multiply the fourth-order Taylor Polynomials for ex and sin x: x3 x4 x3 x2 + + x− 1+x+ 2 6 24 6 = x + x2 − x3 x3 x4 x4 + − + + higher-order terms 6 2 6 6 = x + x2 + x3 + higher-order terms. 3 The terms through degree four in the Maclaurin series for f (x) = ex sin x are therefore x + x2 + x3 . 3 sin x 21. f (x) = = ex ln(1 − x) f (x) 1−x solution Multiply the fourth order Taylor Polynomials for sin x and x3 x− 6 1 + x + x2 + x3 + x4 1 : 1−x = x + x2 − x3 x4 + x3 + x4 − + higher-order terms 6 6 = x + x2 + 5x 4 5x 3 + + higher-order terms. 6 6 The terms through order four of the Maclaurin series for f (x) = x + x2 + sin x are therefore 1−x 5x 4 5x 3 + . 6 6 23. f (x) = (1 + x)1/4 1 f (x) The = first five generalized binomial coefficients for a = 1 are solution 4 1 + sin x 1 −3 1 −3 −7 1 −3 −7 −11 1 3 7 −77 4 4 4 4 4 4 4 4 4 1, , =− , = , = 4 2! 32 3! 128 4! 2048 Therefore, the first four terms in the binomial series for (1 + x)1/4 are 3 1 7 3 77 4 x − x 1 + x − x2 + 4 32 128 2048 25. f (x) = ex tan−1 x −3/2 f (x) = (1 + x) solution Using the Maclaurin series for ex and tan−1 x, we find x3 x3 x3 x4 x4 x2 x3 x −1 + + ··· x− + · · · = x + x2 − + + − + ··· e tan x = 1 + x + 2 6 3 3 2 6 3 1 1 = x + x2 + x3 − x4 + · · · . 6 6 May 23, 2011 700 C H A P T E R 10 INFINITE SERIES 27. f (x) = esin x f (x) = sin (x 3 − x) solution Substituting sin x for x in the Maclaurin series for ex and then using the Maclaurin series for sin x, we find sin3 x sin4 x sin2 x + + + ··· 2 6 24 2 1 x3 1 1 x3 + ··· + x− + ··· + (x − · · · )3 + =1+ x− (x − · · · )4 6 2 6 6 24 esin x = 1 + sin x + 1 1 1 1 1 = 1 + x + x2 − x3 + x3 − x4 + x4 + · · · 2 6 6 6 24 1 1 = 1 + x + x2 − x4 + · · · . 2 8 In Exercises 29–38, find the Taylor series centered at c and find the interval on which the expansion is valid. x f (x) 1= e(e ) 29. f (x) = , c = 1 x solution Write 1 1 = , x 1 + (x − 1) 1 to obtain and then substitute −(x − 1) for x in the Maclaurin series for 1−x ∞ ∞ 1 [−(x − 1)]n = = (−1)n (x − 1)n . x n=0 n=0 This series is valid for |x − 1| < 1. 1 31. f (x) = = e3x,, cc = = 5−1 f (x) 1−x solution Write 1 1 1 1 . = =− · 1−x −4 − (x − 5) 4 1 + x−5 4 1 Substituting − x−5 4 for x in the Maclaurin series for 1−x yields ∞ ∞ n x−5 n n (x − 5) . = = (−1) − n 4 4 1 + x−5 4 n=0 n=0 1 Thus, ∞ ∞ n 1 (x − 5)n 1 n+1 (x − 5) . =− (−1)n = (−1) 1−x 4 4n 4n+1 n=0 n=0 This series is valid for x−5 4 < 1, or |x − 5| < 4. 33. f (x) = x 4 + 3x − 1, cπ= 2 f (x) = sin x, c = solution To determine the2 Taylor series with center c = 2, we compute f (x) = 4x 3 + 3, f (x) = 12x 2 , f (x) = 24x, and f (4) (x) = 24. All derivatives of order five and higher are zero. Now, f (2) = 21, f (2) = 35, f (2) = 48, f (2) = 48, and f (4) (2) = 24. Therefore, the Taylor series is 21 + 35(x − 2) + 48 48 24 (x − 2)2 + (x − 2)3 + (x − 2)4 , 2 6 24 or 21 + 35(x − 2) + 24(x − 2)2 + 8(x − 2)3 + (x − 2)4 . May 23, 2011 Taylor Series S E C T I O N 10.7 1 35. f (x) = =2 ,x 4 c+=3x4− 1, c = 0 f (x) x solution We will first find the Taylor series for x1 and then differentiate to obtain the series for 12 . Write x 1 1 1 1 = = · . x 4 + (x − 4) 4 1 + x−4 4 1 Now substitute − x−4 4 for x in the Maclaurin series for 1−x to obtain ∞ ∞ 1 x−4 n (x − 4)n 1 = (−1)n n+1 . − = x 4 n= 4 4 n=0 Differentiating term-by-term yields ∞ 1 (x − 4)n−1 − 2 = (−1)n n , x 4n+1 n=1 so that ∞ ∞ 1 (x − 4)n−1 (x − 4)n = (−1)n−1 n = (−1)n (n + 1) n+2 . 2 n+1 x 4 4 n=1 n=0 This series is valid for x−4 4 < 1, or |x − 4| < 4. 1√ 37. f (x) = = x, , cc==43 f (x) 1 − x2 solution By partial fraction decomposition 1 1 1 = 2 + 2 , 1−x 1+x 1 − x2 so 1 1 1 1 1 1 1 2 2 + =− · = + · . x−3 2 −2 − (x − 3) 4 + (x − 3) 4 8 1−x 1+ 2 1 + x−3 4 1 Substituting − x−3 2 for x in the Maclaurin series for 1−x gives ∞ ∞ x − 3 n (−1)n = = (x − 3)n , − 2 2n 1 + x−3 2 n=0 n=0 1 while substituting − x−3 4 for x in the same series gives ∞ ∞ x − 3 n (−1)n = = (x − 3)n . − x−3 4 4n 1+ 4 n=0 n=0 1 Thus, ∞ ∞ ∞ ∞ 1 1 (−1)n (−1)n (−1)n+1 (−1)n n+ 1 n= n+ = − (x − 3) (x − 3) (x − 3) (x − 3)n n n 2 n+2 4 2 8 4 1−x 2 22n+3 n=0 n=0 n=0 n=0 ∞ ∞ (−1)n+1 (−1)n+1 (2n+1 − 1) (−1)n = + 2n+3 (x − 3)n = (x − 3)n . n+2 2 2 22n+3 n=0 n=0 This series is valid for |x − 3| < 2. 39. Use the identity1 cos2 x = 12 (1 + cos 2x) to find the Maclaurin series for cos2 x. , c = −1 f (x) = −2 solution The3xMaclaurin series for cos 2x is ∞ n=0 May 23, 2011 (−1)n ∞ (2x)2n 22n x 2n = (−1)n (2n)! (2n)! n=0 701 702 C H A P T E R 10 INFINITE SERIES so the Maclaurin series for cos2 x = 12 (1 + cos 2x) is n 22n x 2n 1+ 1+ ∞ n=1 (−1) (2n)! 2 =1+ ∞ (−1)n n=1 22n−1 x 2n (2n)! 41. Use the Maclaurin series for ln(1 + x) and ln(1 − x) to show that Show that for |x| < 1, 1+x x33 x55 1 ln = x + + + ··· x x 2 tanh 1 −−1x x = x + 3 + 5 + · · · 3 5 for |x| < 1. What can you conclude by comparing this result with that of Exercise 40? 1 d tanh−1 xseries = for 2ln. (1 + x) and ln (1 − x), we have for |x| < 1 Hint: Recall thatthe Maclaurin solution Using dx 1−x ln(1 + x) − ln(1 − x) = ∞ ∞ (−1)n−1 n (−1)n−1 x − (−x)n n n n=1 = n=1 ∞ ∞ ∞ (−1)n−1 n x n 1 + (−1)n−1 n x + = x . n n n n=1 n=1 n=1 Since 1 + (−1)n−1 = 0 for even n and 1 + (−1)n−1 = 2 for odd n, ln (1 + x) − ln (1 − x) = ∞ k=0 2 x 2k+1 . 2k + 1 Thus, ∞ ∞ 1 x 2k+1 1 1 2 1+x ln = (ln(1 + x) − ln(1 − x)) = x 2k+1 = . 2 1−x 2 2 2k + 1 2k + 1 k=0 k=0 Observe that this is the same series we found in Exercise 40; therefore, 1+x 1 ln = tanh−1 x. 2 1−x 1 , that for |x| < 1, 43. Show, by integrating the Maclaurin series for 1 f (x) = 1 2 Maclaurin series of Differentiate the Maclaurin series for twice to 1find . − xthe 1−x (1 − x)3 ∞ 1 · 3 · 5 · · · (2n − 1) x 2n+1 sin−1 x = x + 2 · 4 · 6 · · · (2n) 2n + 1 n=1 solution From Example 10, we know that for |x| < 1 ∞ ∞ 1 · 3 · 5 · · · (2n − 1) 2n 1 · 3 · 5 · · · (2n − 1) 2n 1 x =1+ x , = 2 2 · 4 · 6 · · · (2n) 2 · 4 · 6 · · · (2n) 1−x n=0 n=1 so, for |x| < 1, sin−1 x = ∞ 1 · 3 · 5 · · · (2n − 1) x 2n+1 dx . =C+x+ 2 · 4 · 6 · · · (2n) 2n + 1 1 − x2 n=1 Since sin−1 0 = 0, we find that C = 0. Thus, sin−1 x = x + ∞ 1 · 3 · 5 · · · (2n − 1) x 2n+1 . 2 · 4 · 6 · · · (2n) 2n + 1 n=1 45. How many terms of the Maclaurin series of f (x) = ln(1 + x) are needed to compute ln 1.2 to within an error of at −1 1 Use the first the fivecomputation terms of theand Maclaurin in Exercise 43calculator to approximate most 0.0001? Make compareseries the result with the value. sin 2 . Compare the result with the calculator value. solution Substitute x = 0.2 into the Maclaurin series for ln (1 + x) to obtain: ln 1.2 = ∞ n=1 May 23, 2011 (−1)n−1 ∞ (0.2)n 1 = (−1)n−1 n . n 5 n n=1 S E C T I O N 10.7 This is an alternating series with an = Taylor Series 703 1 . Using the error bound for alternating series n · 5n |ln 1.2 − SN | ≤ aN+1 = 1 , (N + 1)5N+1 so we must choose N so that 1 < 0.0001 (N + 1)5N+1 or (N + 1)5N+1 > 10,000. For N = 3, (N + 1)5N+1 = 4 · 54 = 2500 < 10, 000, and for N = 4, (N + 1)5N+1 = 5 · 55 = 15, 625 > 10, 000; thus, the smallest acceptable value for N is N = 4. The corresponding approximation is: S4 = 4 (−1)n−1 n=1 = 5n · n 1 1 1 1 − 2 + 3 − 4 = 0.182266666. 5 5 ·2 5 ·3 5 ·4 Now, ln 1.2 = 0.182321556, so |ln 1.2 − S4 | = 5.489 × 10−5 < 0.0001. 2 2 47. UseShow the Maclaurin expansion for e−t to express the function F (x) = 0x e−t dt as an alternating power series in x that (Figure 4). 5 π 3 to πapproximate π7 (a) How many terms of the Maclaurin series areπ needed − + − + · the · · integral for x = 1 to within an error of at 3! 5! 7! most 0.001? Carry out the computation check your answer using a computer algebra system. (b) converges to zero. How many termsand must be computed to get within 0.01 of zero? y F(x) T15(x) x 1 2 FIGURE 4 The Maclaurin polynomial T15 (x) for F (t) = solution x e−t dt. 2 0 Substituting −t 2 for t in the Maclaurin series for et yields e−t = 2 ∞ ∞ (−t 2 )n t 2n = ; (−1)n n! n! n=0 n=0 thus, x 0 e−t dt = 2 ∞ (−1)n x 2n+1 . n!(2n + 1) (−1)n 1 . n!(2n + 1) n=0 (a) For x = 1, 1 0 e−t dt = 2 ∞ n=0 1 ; therefore, the error incurred by using SN to approximate the value of This is an alternating series with an = n!(2n+1) the definite integral is bounded by 1 2 1 −t . e dt − SN ≤ aN+1 = 0 (N + 1)!(2N + 3) To guarantee the error is at most 0.001, we must choose N so that 1 < 0.001 (N + 1)!(2N + 3) or (N + 1)!(2N + 3) > 1000. For N = 3, (N + 1)!(2N + 3) = 4! · 9 = 216 < 1000 and for N = 4, (N + 1)!(2N + 3) = 5! · 11 = 1320 > 1000; thus, the smallest acceptable value for N is N = 4. The corresponding approximation is S4 = 4 n=0 May 23, 2011 1 1 1 1 (−1)n =1− + − + = 0.747486772. n!(2n + 1) 3 2! · 5 3! · 7 4! · 9 704 C H A P T E R 10 INFINITE SERIES (b) Using a computer algebra system, we find 1 0 therefore e−t dt = 0.746824133; 2 1 2 −t e dt − S4 = 6.626 × 10−4 < 10−3 . 0 x −4 In Exercises 49–52, express sin tthe dt definite integral as an infinite series and find its value to within an error of at most 10 . Let F (x) = . Show that 1 t 0 cos(x 2 ) dx 49. 0 x3 x5 x7 F (x) = x − + − + ··· 2 3 · 3! 5 ·x5!yields 7 · 7! solution Substituting x for x in the Maclaurin series for cos Evaluate F (1) to three decimal places. ∞ ∞ (x 2 )2n x 4n = ; (−1)n (−1)n cos(x 2 ) = (2n)! (2n)! n=0 n=0 therefore, 1 0 cos(x 2 ) dx = ∞ (−1)n n=0 1 ∞ x 4n+1 (−1)n . = (2n)!(4n + 1) (2n)!(4n + 1) 0 n=0 1 ; therefore, the error incurred by using SN to approximate the value of This is an alternating series with an = (2n)!(4n+1) the definite integral is bounded by 1 1 2 . cos(x ) dx − SN ≤ aN+1 = 0 (2N + 2)!(4N + 5) To guarantee the error is at most 0.0001, we must choose N so that 1 < 0.0001 (2N + 2)!(4N + 5) or (2N + 2)!(4N + 5) > 10,000. For N = 2, (2N + 2)!(4N + 5) = 6! · 13 = 9360 < 10,000 and for N = 3, (2N + 2)!(4N + 5) = 8! · 17 = 685,440 > 10,000; thus, the smallest acceptable value for N is N = 3. The corresponding approximation is S3 = n=0 1 3 −x e 1 dx 51. tan−1 (x 2 ) dx 0 0 solution 3 (−1)n 1 1 1 =1− + − = 0.904522792. (2n)!(4n + 1) 5 · 2! 9 · 4! 13 · 6! Substituting −x 3 for x in the Maclaurin series for ex yields e−x = 3 ∞ ∞ (−x 3 )n x 3n = ; (−1)n n! n! n=0 n=0 therefore, 1 0 3 e−x dx = ∞ (−1)n n=0 1 ∞ x 3n+1 (−1)n . = n!(3n + 1) n!(3n + 1) 0 n=0 1 ; therefore, the error incurred by using SN to approximate the value of This is an alternating series with an = n!(3n+1) the definite integral is bounded by 1 3 1 −x . e dx − SN ≤ aN+1 = 0 (N + 1)!(3N + 4) To guarantee the error is at most 0.0001, we must choose N so that 1 < 0.0001 (N + 1)!(3N + 4) May 23, 2011 or (N + 1)!(3N + 4) > 10,000. S E C T I O N 10.7 Taylor Series 705 For N = 4, (N + 1)!(3N + 4) = 5! · 16 = 1920 < 10,000 and for N = 5, (N + 1)!(3N + 4) = 6! · 19 = 13,680 > 10,000; thus, the smallest acceptable value for N is N = 5. The corresponding approximation is S5 = 5 n=0 (−1)n = 0.807446200. n!(3n + 1) 1 53–56, express the integral as an infinite series. In Exercises dx x 10 − cos(t) x 4 + 1dt, for all x 53. t 0 solution The Maclaurin series for cos t is cos t = ∞ (−1)n n=0 ∞ t 2n t 2n =1+ , (−1)n (2n)! (2n)! n=1 so 1 − cos t = − ∞ ∞ t 2n t 2n = , (−1)n+1 (2n)! (2n)! (−1)n n=1 n=1 and ∞ ∞ 1 t 2n t 2n−1 1 − cos t = (−1)n+1 (−1)n+1 = . t t (2n)! (2n)! n=1 n=1 Thus, x x ∞ ∞ 1 − cos(t) t 2n x 2n n+1 (−1) (−1)n+1 dt = . = t (2n)!2n (2n)!2n 0 n=1 0 n=1 x x 2 ) dt, for |x| < 1 ln(1t + − tsin t dt, for all x 0 t 0 solution Substituting t 2 for t in the Maclaurin series for ln(1 + t) yields 55. ln(1 + t 2 ) = ∞ (−1)n−1 n=1 ∞ (t 2 )n t 2n = . (−1)n n n n=1 Thus, x 0 ln(1 + t 2 ) dt = ∞ n=1 (−1)n x ∞ t 2n+1 x 2n+1 . (−1)n = n(2n + 1) n(2n + 1) 0 n=1 ∞ x dt has Maclaurin series (−1)n 2n x n ? 57. Which function , for |x| < 1 4 n=0 0 1−t solution We recognize that ∞ (−1)n 2n x n = n=0 ∞ (−2x)n n=0 1 with x replaced by −2x. Therefore, is the Maclaurin series for 1−x ∞ (−1)n 2n x n = n=0 1 1 = . 1 − (−2x) 1 + 2x In Exercises 59–62, use has Theorem 2 to prove Which function Maclaurin seriesthat the f (x) is represented by its Maclaurin series for all x. x x ∞ 59. f (x) = sin 2 + cos 3 , (−1)k (x − 3)k ? k+1 solution All derivatives of f(x) consist of sin or cos applied to each of x/2 and x/3 and added together, so each 3 k=0 (n) summand is bounded by 1. Thus f (x) ≤ 2 for all n and x. By Theorem 2, f (x) is represented by its Taylor series for everyFor x. which values of x is the expansion valid? May 23, 2011 706 C H A P T E R 10 INFINITE SERIES 61. f (x) = sinh −x x f (x) = e , solution By definition, sinh x = 21 (ex − e−x ), so if both ex and e−x are represented by their Taylor series centered at c, then so is sinh x. But the previous exercise shows that e−x is so represented, and the text shows that ex is. In Exercises 63–66, find100 the functions with the following Maclaurin series (refer to Table 1 on page 599). f (x) = (1 + x) x9 x 12 x6 63. 1 + x 3 + + + + ··· 2! 3! 4! solution We recognize 1 + x3 + ∞ 3n ∞ x (x 3 )n x9 x 12 x6 + + + ··· = = 2! 3! 4! n! n! n=0 n=0 as the Maclaurin series for ex with x replaced by x 3 . Therefore, 1 + x3 + 3 x9 x 12 x6 + + + · · · = ex . 2! 3! 4! 53 x 3 55 x 5 57 x 7 65. 1 −1 − 4x + − 43 x 3 ++4·4·x·4 − 45 x 5 + · · · + 42 x 2 − 3! 5! 7! solution Note 53 x 3 55 x 5 57 x 7 53 x 3 55 x 5 57 x 7 1− + − + · · · = 1 − 5x + 5x − + − + ··· 3! 5! 7! 3! 5! 7! = 1 − 5x + ∞ (−1)n n=0 (5x)2n+1 . (2n + 1)! The series is the Maclaurin series for sin x with x replaced by 5x, so 1− 55 x 5 57 x 7 53 x 3 + − + · · · = 1 − 5x + sin(5x). 3! 5! 7! In Exercises 6712 and 68,20let 28 x x x + − + ··· x4 − 3 5 7 f (x) = 1 (1 − x)(1 − 2x) 67. Find the Maclaurin series of f (x) using the identity f (x) = solution 1 2 − 1 − 2x 1−x Substituting 2x for x in the Maclaurin series for 1 gives 1−x ∞ ∞ 1 = 2n x n (2x)n = 1 − 2x n=0 n=0 1 is valid for |x| < 1, the two series which is valid for |2x| < 1, or |x| < 12 . Because the Maclaurin series for 1−x together are valid for |x| < 12 . Thus, for |x| < 12 , ∞ ∞ 2 1 1 = − =2 2n x n − xn (1 − 2x)(1 − x) 1 − 2x 1−x n=0 = ∞ n=0 2n+1 x n − ∞ n=0 xn = n=0 ∞ 2n+1 − 1 x n . n=0 69. When a voltage V is applied to a series circuit consisting of a resistor R and an inductor L, the current at time t is Find the Taylor series for f (x) at c = 2. Hint:Rewrite the identity of Exercise 67 as V I (t) = 2 1 − e−Rt/L 1 R − f (x) = −3 − 2(x − 2) −1 − (x − 2) Vt for small t. Expand I (t) in a Maclaurin series. Show that I (t) ≈ L May 23, 2011 S E C T I O N 10.7 solution Taylor Series 707 t Substituting − Rt L for t in the Maclaurin series for e gives n ∞ − Rt ∞ ∞ (−1)n R n n (−1)n R n n L −Rt/L e = = t =1+ t n! n! L n! L n=0 n=0 n=1 Thus, ⎛ 1 − e−Rt/L = 1 − ⎝1 + ∞ (−1)n R n n! n=1 L ⎞ t n⎠ = ∞ (−1)n+1 Rt n n=1 n! L , and ∞ ∞ Vt V (−1)n+1 Rt n V (−1)n+1 Rt n = . + I (t) = R n! L L R n! L n=1 n=2 If t is small, then we can approximate I (t) by the first (linear) term, and ignore terms with higher powers of t; then we find V (t) ≈ Vt . L 3 ) and use it to determine f (6) (0). 71. Find thethe Maclaurin for 69 f (x) cos(x Use result ofseries Exercise and=your knowledge of alternating series to show that solution The Maclaurin series for cos x is R Vt Vt 1− t ≤ I (t) ≤ (for all t) L 2L L ∞ x 2n n cos x = (−1) (2n)! n=0 Substituting x 3 for x gives cos(x 3 ) = ∞ n=0 (−1)n x 6n (2n)! Now, the coefficient of x 6 in this series is − 1 1 f (6) (0) =− = 2! 2 6! so f (6) (0) = − 6! = −360 2 20 73. first of the the Maclaurin Maclaurin series. series for f (x) = ex . How does the result (0) and f (8)to (0)find for the f (x) = three tan−1terms x using Find Use f (7)substitution (k) show that f (0) = 0 for 1 ≤ k ≤ 19? solution Substituting x 20 for x in the Maclaurin series for ex yields ex 20 = ∞ ∞ 20n (x 20 )n x = ; n! n! n=0 n=0 the first three terms in the series are then 1 1 + x 20 + x 40 . 2 (k) Recall that the coefficient of x k in the Maclaurin series for f is f k!(0) . For 1 ≤ k ≤ 19, the coefficient of x k in the 20 Maclaurin series for f (x) = ex is zero; it therefore follows that f (k) (0) =0 k! or f (k) (0) = 0 for 1 ≤ k ≤ 19. Use the binomial series to find f (8) (0) for f (x) = May 23, 2011 1 − x2. 708 C H A P T E R 10 INFINITE SERIES 75. Does the Maclaurin series for f (x) = (1 + x)3/4 converge to f (x) at x = 2? Give numerical evidence to support your answer. solution The Taylor series for f (x) = (1 + x)3/4 converges to f (x) for |x| < 1; because x = 2 is not contained on this interval, the series does not converge to f (x) at x = 2. The graph below displays SN = N 3 4 n n=0 2n for 0 ≤ N ≤ 14. The divergent nature of the sequence of partial sums is clear. SN 15 10 5 0 2 −5 4 6 8 10 12 14 N −10 −15 −20 √ 77. Let f (x) = 1 + x. Explain the steps required to verify that the Maclaurin series for f (x) = ex converges to f (x) for all x. (a) Use a graphing calculator to compare the graph of f with the graphs of the first five Taylor polynomials for f . What do they suggest about the interval of convergence of the Taylor series? (b) Investigate numerically whether or not the Taylor expansion for f is valid for x = 1 and x = −1. solution (a) The five first terms of the Binomial series with a = 12 are 1 1 −1 1 1 −1 1 1 −1 1 −2 1 −2 1 −3 √ 1 2 2 2 2 2 2 2 2 2 1+x =1+ x+ x2 + x3 + x4 + · · · 2 2! 3! 4! 9 45 1 1 = 1 + x − x2 + x3 − x4 + · · · 2 8 4 2 Therefore, the first five Taylor polynomials are T0 (x) = 1; 1 T1 (x) = 1 + x; 2 1 T2 (x) = 1 + x − 2 1 T3 (x) = 1 + x − 2 1 T4 (x) = 1 + x − 2 1 2 x ; 8 1 2 x + 8 1 2 x + 8 1 3 x ; 8 1 3 5 4 x − x . 8 128 The figure displays the graphs of these Taylor polynomials, along with the graph of the function f (x) = is shown in red. 1.5 1 1.5 –1 0.5 0.5 1 The graphs suggest that the interval of convergence for the Taylor series is −1 < x < 1. N 1 2 x n for x = 1 we find (b) Using a computer algebra system to calculate SN = n n=0 S10 = 1.409931183, May 23, 2011 S100 = 1.414073048, S1000 = 1.414209104, √ 1 + x, which S E C T I O N 10.7 which appears to be converging to Taylor Series 709 N 1 √ 2 · (−1)n , and find 2 as expected. At x = −1 we calculate SN = n n=0 S10 = 0.176197052, S100 = 0.056348479, S1000 = 0.017839011, which appears to be converging to zero, though slowly. 79. Use Example 11 and the approximation sin x ≈ x to show that the period T of a pendulum released at an angle θ has Use the first five terms of the Maclaurin series for the elliptic function E(k) to estimate the period T of a 1-meter the following second-order approximation: pendulum released at an angle θ = π4 (see Example 11). θ2 L T ≈ 2π 1+ g 16 solution The period T of a pendulum of length L released from an angle θ is L E(k), T =4 g where g ≈ 9.8 m/s2 is the acceleration due to gravity, E(k) is the elliptic function of the first kind and k = sin θ2 . From Example 11, we know that ∞ π 1 · 3 · 5 · · · (2n − 1) 2 2n E(k) = k . 2 2 · 4 · 6 · · · (2n) n=0 Using the approximation sin x ≈ x, we have k = sin θ θ ≈ ; 2 2 moreover, using the first two terms of the series for E(k), we find " 2 2 # π 1 θ2 θ π E(k) ≈ 1+ 1+ . = 2 2 2 2 16 Therefore, L T =4 E(k) ≈ 2π g L θ2 1+ . g 16 In Exercises 80–83, find the Maclaurin series of the function and use it to calculate the limit. 3 sin x − x + x6 x 2 cos x − 1 + 2 81. lim x→0lim x5 4 x→0 x solution Using the Maclaurin series for sin x, we find sin x = ∞ (−1)n n=0 ∞ x3 x5 x 2n+1 x 2n+1 =x− + + . (−1)n (2n + 1)! 6 120 (2n + 1)! n=3 Thus, ∞ sin x − x + x3 x5 x 2n+1 = (−1)n + 6 120 (2n + 1)! n=3 and 3 sin x − x + x6 x5 ∞ = 1 x 2n−4 + (−1)n 120 (2n + 1)! n=3 Note that the radius of convergence for this series is infinite, and recall from the previous section that a convergent power series is continuous within its radius of convergence. Thus to calculate the limit of this power series as x → 0 it suffices to evaluate it at x = 0: ⎞ ⎛ 3 ∞ 2n−4 sin x − x + x6 x 1 n ⎠= 1 +0= 1 lim + = lim ⎝ (−1) (2n + 1)! 120 120 x→0 x→0 120 x5 n=3 May 23, 2011 710 C H A P T E R 10 INFINITE SERIES 2) cos x sin(x−1 tan x−− x cos x − 16 x 3 83. lim x→0lim x 4 x 25 x→0 x solution We start with sin x = ∞ (−1)n n=0 x 2n+1 (2n + 1)! cos x = ∞ (−1)n n=0 x 2n (2n)! so that ∞ ∞ sin(x 2 ) x 4n+2 x 4n−2 n = (−1) = (−1)n 4 4 (2n + 1)! x (2n + 1)!x n=0 n=0 cos x = x2 ∞ (−1)n n=0 x 2n−2 (2n)! Expanding the first few terms gives ∞ sin(x 2 ) 1 x 4n−2 = 2 − (−1)n 4 (2n + 1)! x x n=1 ∞ cos x 1 1 x 2n−2 = 2 − + (−1)n 2 2 (2n)! x x n=2 so that ∞ ∞ n=1 n=2 4n−2 2n−2 sin(x 2 ) cos x 1 n x nx − − − = (−1) (−1) 2 (2n + 1)! (2n)! x4 x2 Note that all terms under the summation signs have positive powers of x. Now, the radius of convergence of the series for both sin and cos is infinite, so the radius of convergence of this series is infinite. Recall from the previous section that a convergent power series is continuous within its radius of convergence. Thus to calculate the limit of this power series as x → 0 it suffices to evaluate it at x = 0: ⎞ ⎛ ∞ ∞ 4n−2 2n−2 sin(x 2 ) cos x x x 1 ⎠= 1 +0= 1 lim − − 2 (−1)n (−1)n = lim ⎝ − 4 (2n + 1)! (2n)! 2 2 x→0 x→0 2 x x n=1 n=2 Further Insights and Challenges 1 t 85. LetIn g(t) − show2 that . the Maclaurin expansion of f (x) = ln(1 + x) is valid for x = 1. this=exercise we 1 + t2 1+t (a) Show that 1 for all x =π−1, 1 g(t) dt = − ln 2. (a) Show that N 4 2 0 (−1)N+1 x N+1 1 n n 2 3 4 5 = (−1) −xt − t 6 + ... x + (b) Show that g(t) = 1 − t − t + t − t1 + 1+x (c) Evaluate S = 1 − 12 − 13 + 14 − 15 − 16 − 17 +n=0 ... (b) Integrate from 0 to 1 to obtain solution 1 N+1 N (−1)n−1 x dx N+1 + (−1) ln 2 = 1 1 n 1 + x π 1 1 0 n=1 ln(t 2 + 1) = tan−1 1 − ln 2 = − ln 2 g(t) dt = tan−1 t − 2 2 4 2 0 0 (c) Verify that the integral on the right tends to zero as N → ∞ by showing that it is smaller than 01 x N+1 dx. 1 : (b) Start with the series for 1+t (d) Prove the Taylor formula (a) 1∞ 1 1 ln 2 1= 1= − +(−1)−n t n + · · · 2 3 4 1+t n=0 and substitute t 2 for t to get ∞ 1 = (−1)n t 2n = 1 − t 2 + t 4 − t 6 + . . . 2 1+t n=0 May 23, 2011 S E C T I O N 10.7 Taylor Series 711 so that ∞ t = (−1)n t 2n+1 = t − t 3 + t 5 − t 7 + . . . 1 + t2 n=0 Finally, g(t) = 1 t − = 1 − t − t2 + t3 + t4 − t5 − t6 + t7 + . . . 1 + t2 1 + t2 (c) We have 1 1 1 1 1 g(t) dt = (1 − t − t 2 + t 3 + t 4 − t 5 − . . . ) dt = t − t 2 − t 3 + t 4 + t 5 − t 6 − · · · + C 2 3 4 5 6 The radius of convergence of the series for g(t) is 1, so the radius of convergence of this series is also 1. However, this series converges at the right endpoint, t = 1, since 1 1 1 1 1 1− − − + − − ... 2 3 4 5 6 is an alternating series with general term decreasing to zero. Thus by part (a), 1− 1 1 1 1 1 π 1 − + + − − · · · = − ln 2 2 3 4 5 6 4 2 In Exercises 86 and 87, we investigate the convergence of the binomial series ∞ a Ta (x) = n=0 n xn 87. By Exercise 86, Ta (x) converges for |x| < 1, but we do not yet know whether Ta (x) = (1 + x)a . Prove that Ta (x) has radius of convergence R = 1 if a is not a whole number. What is the radius of convergence (a) Verify the identity if a is a whole number? a a a a =n + (n + 1) n n n+1 (b) Use (a) to show that y = Ta (x) satisfies the differential equation (1 + x)y = ay with initial condition y(0) = 1. Ta (x) is zero. (c) Prove that Ta (x) = (1 + x)a for |x| < 1 by showing that the derivative of the ratio (1 + x)a solution (a) a (a − 1) · · · (a − n + 1) a (a − 1) · · · (a − n + 1) (a − n) a a =n· + (n + 1) + (n + 1) · n n+1 n n! (n + 1)! a (a − 1) · · · (a − n + 1) a (a − 1) · · · (a − n + 1) (a − n) + n! (n − 1)! a (a − 1) · · · (a − n + 1) (n + (a − n)) a = =a· n n! = (b) Differentiating Ta (x) term-by-term yields Ta (x) = ∞ a x n−1 . n n n=1 Thus, (1 + x)Ta (x) = ∞ ∞ ∞ ∞ a a a a x n−1 + xn = xn + xn n n (n + 1) n n n n+1 n n=1 = n=1 ∞ $ (n + 1) n=0 a n+1 n=0 +n a n % xn = a n=0 Moreover, Ta (0) = May 23, 2011 n=0 ∞ a 0 = 1. a n x n = aTa (x). 712 C H A P T E R 10 INFINITE SERIES (c) d dx Ta (x) (1 + x)a = (1 + x)Ta (x) − aTa (x) (1 + x)a Ta (x) − a(1 + x)a−1 Ta (x) = = 0. (1 + x)2a (1 + x)a+1 Thus, Ta (x) = C, (1 + x)a for some constant C. For x = 0, Ta (0) 1 = = 1, so C = 1. a (1 + 0) 1 Finally, Ta (x) = (1 + x)a . 2 2 89. Assume that a < b and let π/2 L bethe arc2 length (circumference) of the ellipse xa + yb = 1 shown in Figure 5. 2 that for The function G(k) = 0 1 − k sin t dt is called an elliptic function of the second kind. Prove 2 2 There |k|is<no1,explicit formula for L, but it is known that L = 4bG(k), with G(k) as in Exercise 88 and k = 1 − a /b . Use the first three terms of the expansion of Exercise 88 to estimate L when a = 4 and b = 5. ∞ π π 1 · 3 · · · (2n − 1) 2 k 2n y G(k) = − 2 2 2n − 1 b 2 · · · 4 · (2n) n=1 a FIGURE 5 The ellipse x 2 a x + y 2 b = 1. solution With a = 4 and b = 5, k= and the arc length of the ellipse x 2 4 + y 2 5 42 3 1− 2 = , 5 5 = 1 is ⎛ ⎞ 2 3 2n ∞ π π 1 · 3 · · · − 1) 3 (2n 5 ⎜ ⎟ = 20 ⎝ − L = 20G ⎠. 5 2 2 2 · 4 · · · (2n) 2n − 1 n=1 Using the first three terms in the series for G(k) gives 243 36,157π 1 · 3 2 (3/5)4 9 1 2 (3/5)2 + = 10π 1 − − = ≈ 28.398. · · L ≈ 10π − 10π 2 1 2·4 3 100 40,000 4000 91. Irrationality of e Prove that e is an irrational number using the following argument by contradiction. Suppose that Use Exercise 88 to prove that if a < b and a/b is near 1 (a nearly circular ellipse), then e = M/N, where M, N are nonzero integers. (a) Show that M! e−1 is a whole number. π a2 L ≈ 3b + x B such that M! e−1 equals (b) Use the power series for e at x = −1 to show that there 2 is an integer b Hint: Use the first two terms of the series for G(k). 1 1 − + ··· B + (−1)M+1 M + 1 (M + 1)(M + 2) (c) Use your knowledge of alternating series with decreasing terms to conclude that 0 < |M! e−1 − B| < 1 and observe that this contradicts (a). Hence, e is not equal to M/N . solution Suppose that e = M/N , where M, N are nonzero integers. (a) With e = M/N, M!e−1 = M! which is a whole number. May 23, 2011 N = (M − 1)!N, M Chapter Review Exercises 713 (b) Substituting x = −1 into the Maclaurin series for ex and multiplying the resulting series by M! yields 1 1 (−1)k −1 M!e = M! 1 − 1 + − + · · · + + ··· . 2! 3! k! For all k ≤ M, M! is a whole number, so k! 1 1 (−1)k M! 1 − 1 + − + · · · + 2! 3! M! is an integer. Denote this integer by B. Thus, (−1)M+2 1 (−1)M+1 1 M! e−1 = B + M! + + · · · = B + (−1)M+1 − + ··· . (M + 1)! (M + 2)! M + 1 (M + 1)(M + 2) (c) The series for M! e−1 obtained in part (b) is an alternating series with an = M! n! . Using the error bound for an alternating series and noting that B = SM , we have M! e−1 − B ≤ aM+1 = 1 < 1. M +1 This inequality implies that M! e−1 − B is not a whole number; however, B is a whole number so M! e−1 cannot be a whole number. We get a contradiction to the result in part (a), which proves that the original assumption that e is a rational number is false. Use the result of Exercise 73 in Section 4.5 to show that the Maclaurin series of the function 2 e−1/x for x = 0 f (x) = n−3 0 forinxeach = 0 sequence. 1. Let an = and bn = an+3 . Calculate the first three terms n! whose bn Maclaurin series converges but does not converge (a) aisn2T (x) = 0. This provides an example of a function f (x)(b) to f (x) (except at x = 0). (c) an bn (d) 2an+1 − 3an CHAPTER REVIEW EXERCISES solution (a) 1−3 2 = (−2)2 = 4; 1! 2−3 2 1 2 1 = = − = ; 2! 2 4 3−3 2 = = 0. 3! a12 = a22 a32 (b) 4−3 1 = ; 4! 24 1 5−3 = ; b2 = a5 = 5! 60 1 6−3 = . b3 = a6 = 6! 240 b1 = a4 = (c) Using the formula for an and the values in (b) we obtain: 1−3 1! 2−3 a2 b2 = 2! 3−3 a3 b3 = 3! a1 b1 = May 23, 2011 1 1 =− ; 24 12 1 1 · =− ; 60 120 1 · = 0. 240 · 714 C H A P T E R 10 INFINITE SERIES (d) 1 − 3(−2) = 5; 2a2 − 3a1 = 2 − 2 3 1 = ; 2a3 − 3a2 = 2 · 0 − 3 − 2 2 2a4 − 3a3 = 2 · 1 1 −3·0= . 24 12 In Exercises 3–8, compute2n the−limit lim an = 2. 2 state that it does not exist) assuming that n→∞ 1 (or Prove that lim = using the limit definition. n→∞ 3n + 2 3 3. lim (5an − 2an2 ) n→∞ solution 2 5an − 2an2 = 5 lim an − 2 lim an2 = 5 lim an − 2 lim an = 5 · 2 − 2 · 22 = 2. lim n→∞ n→∞ n→∞ n→∞ n→∞ 5. lim ean n→∞ 1 lim n→∞ a solution Thenfunction f (x) = ex is continuous, hence: lim ean = elimn→∞ an = e2 . n→∞ 7. lim (−1)n an n→∞ lim cos(πan ) n→∞ solution Because lim an = 0, it follows that lim (−1)n an does not exist. n→∞ n→∞ In Exercises 9–22, determine the limit of the sequence or show that the sequence diverges. an + n lim √ √ 2 n→∞ 9. an = n a+n 5+−n n + 2 solution First rewrite an as follows: √ √ √ √ n+5− n+2 n+5+ n+2 (n + 5) − (n + 2) 3 = √ = √ . an = √ √ √ √ n+5+ n+2 n+5+ n+2 n+5+ n+2 Thus, 3 lim an = lim √ = 0. √ n→∞ n + 5 + n + 2 n→∞ 2 11. an = 21/n 3 3n − n an =The function x solution 1 − 2n3 f (x) = 2 is continuous, so 2 2 lim an = lim 21/n = 2limn→∞ (1/n ) = 20 = 1. n→∞ n→∞ m 13. bm = 1 + (−1) 10n an =Because 1 + (−1)m is equal to 0 for m odd and is equal to 2 for m even, the sequence {bm } does not approach solution n! one limit; hence this sequence diverges. n+2 15. bn = tan−1 1 + (−1)m n+5 bm = m solution The function tan−1 x is continuous, so π n+2 n+2 lim bn = lim tan−1 = tan−1 lim = tan−1 1 = . n→∞ n→∞ n→∞ n + 5 n+5 4 2 17. bn = n2100 + nn − 3 n + π+n 1 − an = solution Rewrite n! bn as5n n2 + n − n2 + 1 n2 + n + n2 + 1 n2 + n − n2 + 1 n−1 . bn = = = 2 2 2 2 2 n + n + n2 + 1 n +n+ n +1 n +n+ n +1 May 23, 2011 Chapter Review Exercises Then lim bn = lim n→∞ n→∞ n − 1 1 − n1 1−0 1 n n = lim = √ = . √ 2 2 n→∞ 2 n + n + n + 1 1+0+ 1+0 1 + n1 + 1 + 12 n2 n2 n2 n2 n 1 3m 2 19. bm c=n =1 +n + n − n2 − n m 1 m solution lim bm = lim 1 + = e. m→∞ m→∞ m 21. bn = n ln(n ln n + 1)− 3 n 1+ cn =Write solution n bn = n ln n+1 n = ln 1 + n1 1 n . Using L’Hôpital’s Rule, we find −1 ln 1 + n1 ln 1 + x1 · − 12 1 + x1 1 −1 x lim bn = lim = lim = lim = lim = 1. 1 + 1 1 n→∞ n→∞ x→∞ x→∞ x→∞ x − 12 n x x 23. Use the Squeeze to show that lim ln(n2 +Theorem 1) n→∞ cn = ln(n3 + 1) solution For all x, − arctan(n2 ) = 0. √ n π π < arctan x < , 2 2 so π/2 arctan(n2 ) π/2 −√ < < √ , √ n n n for all n. Because π/2 −√ n→∞ n lim π/2 = lim √ = 0, n→∞ n it follows by the Squeeze Theorem that arctan(n2 ) = 0. √ n→∞ n lim an+1 1 n 1 n 3 − {a 2 . } such that {sin an } is convergent. an = sequence 25. Calculate Give anlim example of, where a divergent n→∞ an 2 3 n solution Because 3n 1 n 1 n 1 n 1 n 3 − 2 ≥ 3 − 3 = 2 3 2 3 6 and lim 3n n→∞ 6 = ∞, we conclude that limn→∞ an = ∞, so L’Hôpital’s rule may be used: 1 3n+1 − a lim n+1 = lim 2 1 n n→∞ an n→∞ 23 − n+1 1 2n+1 n+2 − 2n+2 3 − 2 23 3 3−0 3 = 3. = lim n+1 = lim n+1 = 1 2n n+1 n→∞ 3 n→∞ 1 −0 −2 3 1 − 23 ∞ n−2 = sums an + S64with Definethe an+1 27. Calculate partial and aS17 = of 2. the series . n2 + 2n n=1 (a) Compute an for n = 2, 3, 4, 5. solution (b) Show that {an } is increasing and is bounded by 3. 1 2 11 1 + 0 find + its+value.= − = −0.183333; − and (c) Prove that lim Sa4n = exists n→∞ 3 15 24 60 1 2 3 4 5 287 1 + + + + = = 0.065079. S7 = − + 0 + 3 15 24 35 48 63 4410 May 23, 2011 715 716 C H A P T E R 10 INFINITE SERIES 8 16 32 4 29. Find the sum + 1+ 1 + 1 + · · · . 9 1− 27 +81 −243 + · · · . Find the sum 4 42 43 solution This is a geometric series with common ratio r = 23 . Therefore, 4 4 4 8 16 32 + + + + ··· = 9 2 = . 9 27 81 243 3 1− 3 ∞ n+3 ∞2 n 31. Find the sum 2. 3n . Find the sum n=−1 e n=2 solution Note ∞ ∞ n ∞ 2n+3 2n 2 3 = 2 = 8 ; n n 3 3 3 n=−1 therefore, n=−1 n=−1 ∞ 1 3 2n+3 =8· · = 36. 3n 2 1− 2 3 n=−1 ∞ ∞ ∞ ∞ divergent an and π bn such that (an + bn ) = 1. 33. Give an example of series −1 2 b − tan n diverges if b = . Show that n=1 n=1 n=1 2 n=1 n solution Let an = 12 + 1, bn = −1. The corresponding series diverge by the Divergence Test; however, ∞ (an + bn ) = n=1 ∞ n 1 n=1 2 = 1 2 = 1. 1 − 21 ∞ 1 ∞ 35. Evaluate S = 1 1 . n(n + 3) − . Compute SN for N Let S = n=3 n n+2 n=1 solution Note that 3 1 SN = = n(n + 3) 2 = 1, 2, 3, 4. Find S by showing that 1 1 1 11 − − − 3 Nn+ 1n +N3 + 2 so that N n=3 N 1 1 1 1 = − n(n + 3) 3 n n+3 n=3 1 1 1 1 1 1 − + − + − 3 6 4 7 5 8 1 1 1 1 1 1 − + ··· + − + − 6 9 N −1 N +2 N N +3 1 1 1 1 1 1 1 + + − − − = 3 3 4 5 N +1 N +2 N +3 = 1 3 Thus ∞ n=3 N 1 1 = lim n(n + 3) 3 N→∞ 1 = 3 n=3 1 1 − n n+3 1 1 1 1 1 1 + + − − − 3 4 5 N +1 N +2 N +3 1 = 3 1 1 1 + + 3 4 5 In Exercises usearea the of Integral Test to many determine whether infinite converges. Find37–40, the total the infinitely circles on thethe interval [0,series 1] in Figure 1. 37. ∞ n2 n=1 n3 + 1 solution 2 Let f (x) = 3x . This function is continuous and positive for x ≥ 1. Because x +1 f (x) = May 23, 2011 (x 3 + 1)(2x) − x 2 (3x 2 ) x(2 − x 3 ) = , (x 3 + 1)2 (x 3 + 1)2 = 47 180 Chapter Review Exercises 717 we see that f (x) < 0 and f is decreasing on the interval x ≥ 2. Therefore, the Integral Test applies on the interval x ≥ 2. Now, R ∞ x2 x2 1 3 + 1) − ln 9 = ∞. ln(R lim dx = lim dx = 3 R→∞ R→∞ 2 x 3 + 1 2 x3 + 1 The integral diverges; hence, the series ∞ n=2 n=1 ∞ 1 ∞ n2 (n + 2)(ln(n + 2))3 n=1 (n3 + 1)1.01 n=1 1 solution Let f (x) = 3 39. ∞ n2 n2 diverges, as does the series . n3 + 1 n3 + 1 (x+2) ln (x+2) ∞ 0 1 dx, we have . Using the substitution u = ln(x + 2), so that du = x+2 ∞ ∞ 1 1 1 R f (x) dx = du = lim du = lim − 2 R→∞ ln 2 u3 R→∞ 2u ln 2 ln 2 u3 1 1 1 − = = lim R→∞ 2(ln 2)2 2(ln R)2 2(ln 2)2 Since the integral of f (x) converges, so does the series. In Exercises use the Comparison or Limit Comparison Test to determine whether the infinite series converges. ∞ 41–48, n3 ∞ 1n4 n=1 e 41. (n + 1)2 n=1 For all n ≥ 1, solution 0< The series 1 1 < n+1 n so 1 1 < 2. 2 (n + 1) n ∞ ∞ 1 1 is a convergent p-series, so the series converges by the Comparison Test. n2 (n + 1)2 n=1 n=1 ∞ 2 ∞n + 1 43. 1 n3.5√− 2 n=2 n+n n=1 2 1 . Now, solution Apply the Limit Comparison Test with an = n3.5+1 and bn = 1.5 n −2 n n2 +1 3.5 n3.5 + n1.5 L = lim n 1−2 = lim = 1. n→∞ n→∞ n3.5 − 2 n1.5 Because L exists and ∞ 1 is a convergent p-series, we conclude by the Limit Comparison Test that the series n1.5 n=1 ∞ n2 + 1 also converges. n3.5 − 2 n=2 ∞ ∞ n 1 5 n=2 nn + − 5ln n n=1 solution For all n ≥ 2, 45. The series ∞ n=2 47. n n 1 < 5/2 = 3/2 . n n n5 + 5 1 is a convergent p-series, so the series n3/2 ∞ n converges by the Comparison Test. 5 n=2 n + 5 ∞ n10 + 10n ∞ 1 n11 + 11n n=1 3n − 2n n n=1 10 n solution Apply the Limit Comparison Test with an = n11 +10n and bn = 10 11 . Then, n +11 n10 +10n n n +10 n +1 11 n an n n = lim n +11 L = lim = 1. n = lim 1110 n = lim 1011 n→∞ bn n→∞ n→∞ n→∞ n +11 n 10 11n 11n + 1 11 May 23, 2011 10 10 718 C H A P T E R 10 INFINITE SERIES The series ∞ n 10 11 n=1 is a convergent geometric series; because L exists, we may therefore conclude by the Limit Comparison Test that the series ∞ 10 n + 10n also converges. n11 + 11n n=1 ∞ ∞ 20 n 49. Determine convergence of n the + 21 n=1 n21 + 20n n=1 n +n 2 solution With an = 3n −2 , we have n 2n + n using the Limit Comparison Test with bn = 23 . n 3 −2 n 1 n n n n 1 + n an 2 +n 3 6 + n3 2 L = lim · n = lim n = lim n = lim n = 1 n→∞ bn n→∞ 3 − 2 2 n→∞ 6 − 2n+1 n→∞ 1 1−2 3 Since L = 1, the two series either both converge or both diverge. Since Limit Comparison Test tells us that ∞ n 2 +n also converges. 3n − 2 ∞ n 2 n=1 3 is a convergent geometric series, the n=1 ∞ ∞ 1 . lim lnan an diverges. Hint: Show that a1n ≥ 2n 51. Let an = 1 − 1 − n1 . Show that n = 0 and that . Determine the convergence of n→∞ n using the Limit Comparison Test with bn = n n=1 1.5 1.4 n=1 solution √ √ n−1 1 n− n−1 n − (n − 1) 1 1− 1− =1− = = √ √ = √ √ n n n n( n + n − 1) n + n2 − n ≥ The series n+ 1 √ n2 = 1 . 2n ∞ 1 1 also diverges by the Comparison Test. diverges, so the series ∞ 1 − 1 − n=2 n 2n n=2 ∞ n ∞ . 53. Let S = 1 2 + 1)2 1 − 1 − 2 converges. Determine(nwhether n=1 n (a) Show that S converges. n=2 Use Eq. (4) in Exercise 83 of Section 10.3 with M = 99 to approximate S. What is the maximum size of the (b) error? solution (a) For n ≥ 1, The series n 1 n < 2 2 = 3. (n2 + 1)2 (n ) n ∞ ∞ 1 n is a convergent p-series, so the series also converges by the Comparison Test. n3 (n2 + 1)2 n=1 n=1 (b) With an = 2 n 2 , f (x) = 2 x 2 and M = 99, Eq. (4) in Exercise 83 of Section 10.3 becomes (n +1) (x +1) ∞ ∞ 100 n x n x + dx ≤ S ≤ + dx, 2 2 2 2 2 2 2 (n + 1) (n + 1) 100 (x + 1) 100 (x + 1)2 n=1 n=1 99 ⎞ ∞ n x 100 + dx ⎠ ≤ . 0≤S−⎝ (n2 + 1)2 (x 2 + 1)2 (1002 + 1)2 100 n=1 ⎛ or 99 Now, 99 n=1 n = 0.397066274; and (n2 + 1)2 R 1 x x 1 1 dx = lim dx = lim − + 2 R→∞ R→∞ 100 (x 2 + 1)2 R 2 + 1 1002 + 1 100 (x 2 + 1)2 ∞ = May 23, 2011 1 = 0.000049995; 20002 Chapter Review Exercises 719 thus, S ≈ 0.397066274 + 0.000049995 = 0.397116269. The bound on the error in this approximation is 100 = 9.998 × 10−7 . (1002 + 1)2 In Exercises 54–57, determine whether the series converges absolutely. If it does not, determine whether it converges conditionally. ∞ n ∞ (−1) n (−1) 1.1 n √ ln(n + 1) 3 n=1 n + 2n 55. n=1 solution Consider the corresponding positive series ∞ n=1 1 . Because n1.1 ln(n + 1) 1 n1.1 ln(n + 1) and 1 < 1.1 n ∞ ∞ 1 (−1)n is a convergent p-series, we can conclude by the Comparison Test that also converges. n1.1 n1.1 ln(n + 1) n=1 ∞ Thus, n=1 n=1 (−1)n converges absolutely. 1.1 n ln(n + 1) ∞ cos π4 +π 2πn ∞ cos√ 4 + πn n√ n=1 n n=1 √ π solution cos 4 + 2πn = cos π4 = 22 , so 57. √ ∞ ∞ cos π4 + 2πn 2 1 = √ √ . 2 n n n=1 n=1 π This is a divergent p-series, so the series ∞ cos 4 + 2πn diverges. √ n n=1 ∞ (−1)k ∞ . 59. Catalan’s constant is defined by K = (−1)n Use a computer algebra system to approximate (2k + 1)2 √ to within an error of at most 10−5 . 3 k=0 n + n n=1 (a) How many terms of the series are needed to calculate K with an error of less than 10−6 ? (b) Carry out the calculation. solution Using the error bound for an alternating series, we have |SN − K| ≤ 1 1 = . (2(N + 1) + 1)2 (2N + 3)2 For accuracy to three decimal places, we must choose N so that 1 < 5 × 10−3 (2N + 3)2 or (2N + 3)2 > 2000. Solving for N yields N> 1 √ 2000 − 3 ≈ 20.9. 2 Thus, K≈ 21 k=0 May 23, 2011 (−1)k = 0.915707728. (2k + 1)2 720 C H A P T E R 10 INFINITE SERIES ∞ ∞ ∞ ∞ an be an absolutely convergent series. Determine whether the following series are convergent or divergent: Give an example of conditionally convergent series an and bn such that (an + bn ) converges abson=1 n=1 ∞ n=1 n=1 ∞ 1 (a) lutely. (−1)n an an + 2 (b) n 61. Let (c) n=1 ∞ 1 (d) 1 + an2 n=1 solution n=1 Because ∞ an converges absolutely, we know that n=1 (a) Because we know that ∞ n=1 1 an + 2 n ∞ an converges and the series n=1 n=1 (c) Because an converges and that ∞ |an | converges. n=1 ∞ 1 is a convergent p-series, the sum of these two series, n2 n=1 also converges. ∞ ∞ (−1)n an = |an | n=1 ∞ ∞ |an | n n=1 (b) We have, Because n=1 ∞ |an | converges, it follows that ∞ ∞ n=1 (−1)n an converges absolutely, which implies that n=1 ∞ (−1)n an converges. n=1 an converges, limn→∞ an = 0. Therefore, n=1 lim 1 n→∞ 1 + a 2 n and the series ∞ 1 1 + an2 n=1 = 1 = 1 = 0, 1 + 02 diverges by the Divergence Test. ∞ ∞ |an | |an | converges, so the series also converges by the Comparison Test. (d) |ann | ≤ |an | and the series n n=1 n=1 √ In Exercises 63–70, apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive. Let {an } be a positive sequence such that lim n an = 12 . Determine whether the following series converge or n→∞ ∞ n5 diverge: 63. n ∞ ∞ 5∞ √ n=1 2an (b) 3n an (c) an (a) 5 solutionn=1With an = n5n , n=1 n=1 an+1 (n + 1)5 5n 1 5 1 = 1 + · = , a 5 n n5 5n+1 n and an+1 1 1 1 1 5 = ρ = lim lim 1 + = ·1= . n→∞ an 5 n→∞ n 5 5 Because ρ < 1, the series converges by the Ratio Test. ∞ 1 ∞ √ n+1 n n2 + n83 n=1 n n=1 solution With an = 65. 1 , n2n +n3 2 2 n + n3 n2n 1 + n2n an+1 1 + n2n 1 n n2 = = · · , a (n + 1)2n+1 + (n + 1)3 = 2 2 n + 1 1 + (n+1)2 n (n + 1)2n+1 1 + (n+1) n+1 n+1 2 and a 1 1 ρ = lim n+1 = · 1 · 1 = . n→∞ an 2 2 Because ρ < 1, the series converges by the Ratio Test. May 23, 2011 2 Chapter Review Exercises 67. 721 ∞ n2 2 4 ∞ n n! n=1 n! n=1 n2 solution With an = 2n! , (n+1)2 an+1 n! 22n+1 = 2 a (n + 1)! · n2 = n + 1 n 2 a ρ = lim n+1 = ∞. n→∞ an and Because ρ > 1, the series diverges by the Ratio Test. ∞ ∞n n 1 ln n 2 n! n=1 n3/2 n=4 n 1 , solution With an = n2 n! n+1 n an+1 1 2 1 n 1 n+1 n 1 = n+1 · 1 + n! = = , a 2 (n + 1)! n 2 n 2 n n 69. and a 1 ρ = lim n+1 = e. n→∞ an 2 Because ρ = 2e > 1, the series diverges by the Ratio Test. In Exercises the Root Test to determine convergence or divergence, or state that the Root Test is inconclusive. ∞ 71–74, n n apply 1 ∞ n! 1 4 n=1 71. 4n n=1 solution With an = 41n , L = lim n→∞ √ n a = lim n n→∞ n 1 1 = . 4n 4 Because L < 1, the series converges by the Root Test. n ∞ ∞ 3 n 73. 2 4n n=1 n n=1 n 3 solution With an = 4n , L = lim √ n n→∞ an = lim n n→∞ 3 n 3 = lim = 0. n→∞ 4n 4n Because L < 1, the series converges by the Root Test. In Exercises determine convergence or divergence using any method covered in the text. 3 ∞ 75–92, 1 n cos ∞ 2 n n n=1 75. 3 n=1 solution This is a geometric series with ratio r = 23 < 1; hence, the series converges. 77. ∞ ∞−0.02n e π 7n n=1 e8n n=1 solution This is a geometric series with common ratio r = 79. 1 ≈ 0.98 < 1; hence, the series converges. e0.02 ∞ n−1 ∞ (−1) √ √ −0.02n nne + n+1 n=1 n=1 solution In this alternating series, an = √ 1√ . The sequence {an } is decreasing, and n+ n+1 lim an = 0; n→∞ therefore the series converges by the Leibniz Test. May 23, 2011 722 C H A P T E R 10 INFINITE SERIES 81. ∞ n (−1) ∞ 1 ln n n=2 n(ln n)3/2 n=10 solution The sequence an = ln1n is decreasing for n ≥ 10 and lim an = 0; n→∞ therefore, the series converges by the Leibniz Test. 83. ∞ ∞ n1 √e n n + ln n n=1 n! n=1 solution For n ≥ 1, 1 1 1 ≤ √ = 3/2 . √ n n n n n + ln n The series ∞ n=1 85. 1 is a convergent p-series, so the series n3/2 ∞ ∞ 1 1 √ −√ 1 n3 √ √ n +1 n=1 n(1 + n) ∞ 1 converges by the Comparison Test. √ n n + ln n n=1 n=1 solution This series telescopes: ∞ 1 1 1 1 1 1 1 + √ −√ + √ −√ + ... = 1− √ √ −√ n n+1 2 2 3 3 4 n=1 so that the nth partial sum Sn is 1 1 1 1 1 1 1 1 Sn = 1 − √ + √ −√ + √ −√ + ··· + √ − √ =1− √ n n+1 n+1 2 2 3 3 4 and then ∞ 1 1 1 = lim Sn = 1 − lim √ =1 √ −√ n→∞ n→∞ n n+1 n+1 n=1 87. ∞ ∞ 1 √ n + ln nn − ln(n + 1) n=1 n=1 solution For n ≥ 1, √ n ≤ n, so that ∞ n=1 ∞ 1 1 √ ≥ 2n n+ n n=1 which diverges since it is a constant multiple of the harmonic series. Thus ∞ n=1 1 √ diverges as well, by the Comparison n+ n Test. ∞ ∞1 cos(πn) nln n 2/3 n=2 n n=2 solution For n ≥ N large enough, ln n ≥ 2 so that 89. ∞ n=N 1 nln n ≤ n=N which is a convergent p-series. Thus by the Comparison Test, n < N does not affect convergence. ∞ 1 ln3 n n=2 May 23, 2011 ∞ 1 n2 ∞ n=N 1 nln n also converges; adding back in the terms for Chapter Review Exercises 91. ∞ sin2 n=1 π n For all x > 0, sin x < x. Therefore, sin2 x < x 2 , and for x = πn , solution sin2 The series 723 1 π π2 < 2 = π2 · 2 . n n n ∞ ∞ 1 π is a convergent p-series, so the series sin2 also converges by the Comparison Test. 2 n n n=1 n=1 In Exercises find the interval of convergence of the power series. ∞ 93–98, 22n ∞ n n 2 x n! n=0 93. n! n=0 n n solution With an = 2 n!x , 2n+1 x n+1 a 2 n! · n n = lim x · = 0 ρ = lim n+1 = lim n→∞ an n→∞ (n + 1)! 2 x n→∞ n Then ρ < 1 for all x, so that the radius of convergence is R = ∞, and the series converges for all x. 95. ∞ 6 ∞n n x (x − 3)n 8 n +1 n=0 n+1 n=0 6 n solution With an = n (x−3) , n8 +1 (n + 1)6 (x − 3)n+1 an+1 n8 + 1 ρ = lim = lim · 6 n→∞ an n→∞ (n + 1)8 − 1 n (x − 3)n (n + 1)6 (n8 + 1) = lim (x − 3) · 6 n→∞ n ((n + 1)8 + 1) n14 + terms of lower degree = lim (x − 3) · 14 = |x − 3| n→∞ n + terms of lower degree Then ρ < 1 when |x − 3| < 1, so the radius of convergence is 1, and the series converges absolutely for |x − 3| < 1, or ∞ n6 , which converges by the Comparison Test comparing 2 < x < 4. For the endpoint x = 4, the series becomes 8 n +1 n=0 ∞ ∞ 6 1 n (−1)n . For the endpoint x = 2, the series becomes , which converges by the with the convergent p-series n2 n8 + 1 n=1 ∞ 6 n (x − 3)n therefore converges for 2 ≤ x ≤ 4. Leibniz Test. The series n8 + 1 n=0 n=0 97. ∞ ∞ (nx)n n nx n=0 n=0 solution With an = nn x n , and assuming x = 0, n (n + 1)n+1 x n+1 an+1 = lim x(n + 1) · n + 1 = ∞ = lim ρ = lim n n n→∞ an n→∞ n→∞ n x n n 1 n converges to e and the (n + 1) term diverges to ∞. Thus ρ < 1 only when x = 0, so the = 1 + since n+1 n n series converges only for x = 0. 2 ∞ f (x) = n 99. Expand (2x − 3)4 − 3x as a power series centered at c = 0. Determine the values of x for which the series converges. n ln n solution Write n=0 1 1 2 = . 4 − 3x 2 1 − 3x 4 May 23, 2011 724 C H A P T E R 10 INFINITE SERIES 1 , we obtain Substituting 34 x for x in the Maclaurin series for 1−x 1 1 − 34 x = ∞ n 3 n=0 4 xn. This series converges for 34 x < 1, or |x| < 34 . Hence, for |x| < 43 , ∞ 2 1 3 n n = x . 4 − 3x 2 4 n=0 ∞ x 2k . 101. LetProve F (x) that = 2k · k! ∞ k=0 (a) Show that F (x) has infinite radius of convergence.ne−nx = n=0 (b) Show that y = F (x) is a solution of e−x (1 − e−x )2 Hint: Express the left-hand side as the derivative of a geometric series. y(0) = 1, y (0) = 0 y = xy + y, Plot the partial sums SN for N = 1, 3, 5, 7 on the same set of axes. (c) solution 2k (a) With ak = xk , 2 ·k! ak+1 x2 |x|2k+2 2k · k! a = 2k+1 · (k + 1)! · |x|2k = 2(k + 1) , k and ak+1 = x 2 · 0 = 0. ρ = lim k→∞ ak Because ρ < 1 for all x, we conclude that the series converges for all x; that is, R = ∞. (b) Let y = F (x) = ∞ x 2k . 2k · k! k=0 Then y = ∞ ∞ 2kx 2k−1 x 2k−1 = , 2k k! 2k−1 (k − 1)! k=1 k=1 ∞ (2k − 1)x 2k−2 y = , 2k−1 (k − 1)! k=1 and xy + y = x ∞ k=1 ∞ ∞ ∞ x 2k−1 x 2k x 2k x 2k = + + 1 + 2k k! 2k k! 2k−1 (k − 1)! 2k−1 (k − 1)! k=0 k=1 k=1 ∞ ∞ ∞ (2k + 1)x 2k (2k + 1)x 2k (2k − 1)x 2k−2 = = = y . =1+ 2k k! 2k k! 2k−1 (k − 1)! k=1 k=0 k=1 Moreover, y(0) = 1 + ∞ 02k =1 2k k! k=1 Thus, and y (0) = ∞ k=1 02k−1 = 0. 2k−1 (k − 1)! ∞ x 2k is the solution to the equation y = xy + y satisfying y(0) = 1, y (0) = 0. 2k k! k=0 May 23, 2011 Chapter Review Exercises (c) The partial sums S1 , S3 , S5 and S7 are plotted in the figure below. y 7 6 5 4 3 2 1 −2 −1 1 2 x In Exercises 103–112, find the Taylor series centered at c. ∞ Find a power series P (x) = an x n that satisfies the Laguerre differential equation 103. f (x) = e4x , c = 0 n=0 Substituting 4x for x in the Maclaurin series for ex yields xy + (1 − x)y − y = 0 ∞ ∞ n (4x)n 4 n = x . with initial condition satisfying P (0) =e4x 1. = n! n! solution n=0 n=0 105. f (x) = x 4 , 2xc = 2 f (x) = e , c = −1 solution We have f (x) = 4x 3 f (x) = 12x 2 f (x) = 24x f (4) (x) = 24 and all higher derivatives are zero, so that f (2) = 24 = 16 f (2) = 4 · 23 = 32 f (2) = 12 · 22 = 48 f (2) = 24 · 2 = 48 f (4) (2) = 24 Thus the Taylor series centered at c = 2 is 4 48 32 48 24 f (n) (2) (x − 2)n = 16 + (x − 2) + (x − 2)2 + (x − 2)3 + (x − 2)4 n! 1! 2! 3! 4! n=0 = 16 + 32(x − 2) + 24(x − 2)2 + 8(x − 2)3 + (x − 2)4 107. f (x) = sin x, c = π f (x) = x 3 − x, c = −2 solution We have f (4n) (x) = sin x f (4n+1) (x) = cos x f (4n+2) (x) = − sin x f (4n+3) (x) = − cos x so that f (4n) (π) = sin π = 0 f (4n+1) (π) = cos π = −1 f (4n+2) (π) = − sin π = 0 f (4n+3) (π ) = − cos π = 1 Then the Taylor series centered at c = π is ∞ 1 −1 −1 1 f (n) (π) (x − π)n = (x − π) + (x − π)3 + (x − π)5 + (x − π )7 − . . . n! 1! 3! 5! 7! n=0 1 1 1 (x − π)5 + (x − π )7 − . . . = −(x − π) + (x − π)3 − 6 120 5040 1 109. f (x) = = ex−1, , cc==−2 −1 f (x) 1 − 2x solution Write 1 1 1 1 = = . 1 − 2x 5 − 2(x + 2) 5 1 − 2 (x + 2) 5 1 yields Substituting 25 (x + 2) for x in the Maclaurin series for 1−x 1 1 − 25 (x + 2) May 23, 2011 = ∞ n 2 (x + 2)n ; 5n n=0 725 726 C H A P T E R 10 INFINITE SERIES hence, ∞ ∞ 1 1 2n 2n n= = (x + 2) (x + 2)n . 1 − 2x 5 5n 5n+1 n=0 x 111. f (x) = ln , c1 = 2 f (x) = 2 , 2 solution Write(1 − 2x) n=0 c = −2 x (x − 2) + 2 x−2 = ln = ln 1 + . 2 2 2 ln Substituting x−2 2 for x in the Maclaurin series for ln(1 + x) yields ln n ∞ (−1)n+1 x−2 2 x = 2 n=1 n = ∞ (−1)n+1 (x − 2)n . n · 2n n=1 This series is valid for |x − 2| < 2. x first three terms of the Maclaurin series of f (x) and use it to calculate f (3) (0). In Exercises 113–116,find the , c=0 f (x) = x ln 1 + 2 2 2 x 113. f (x) = (x − x)e solution Substitute x 2 for x in the Maclaurin series for ex to get 2 1 1 ex = 1 + x 2 + x 4 + x 6 + . . . 2 6 so that the Maclaurin series for f (x) is 2 1 1 (x 2 − x)ex = x 2 + x 4 + x 6 + · · · − x − x 3 − x 5 − · · · = −x + x 2 − x 3 + x 4 + . . . 2 2 The coefficient of x 3 is f (0) = −1 3! so that f (0) = −6. 1 115. f (x) = = tan−1 (x 2 − x) f (x) 1 + tan x solution 1 to get Substitute − tan x in the Maclaurin series for 1−x 1 = 1 − tan x + (tan x)2 − (tan x)3 + . . . 1 + tan x We have not yet encountered the Maclaurin series for tan x. We need only the terms up through x 3 , so compute tan (x) = sec2 x tan (x) = 2(tan x) sec2 x tan (x) = 2(1 + tan2 x) sec2 x + 4(tan2 x) sec2 x so that tan (0) = 1 tan (0) = 0 tan (0) = 2 Then the Maclaurin series for tan x is tan x = tan 0 + tan (0) 2 tan (0) 3 1 tan (0) x+ x + x + · · · = x + x3 + . . . 1! 2! 3! 3 Substitute these into the series above to get 2 3 1 1 1 1 = 1 − x + x3 + x + x3 − x + x3 + . . . 1 + tan x 3 3 3 1 = 1 − x − x 3 + x 2 − x 3 + higher degree terms 3 4 = 1 − x + x 2 − x 3 + higher degree terms 3 May 23, 2011 Chapter Review Exercises The coefficient of x 3 is 4 f (0) =− 3! 3 so that f (0) = −6 · 4 = −8 3 π π7 π 3√ π5 117. Calculate − x)3 1++ x5 − 7 + · · · . f (x) = (sin 2 2 3! 2 5! 2 7! solution We recognize that ∞ π π7 π3 (π/2)2n+1 π5 − 3 + 5 − 7 + ··· = (−1)n 2 (2n + 1)! 2 3! 2 5! 2 7! n=0 is the Maclaurin series for sin x with x replaced by π/2. Therefore, π7 π π π3 π5 − 3 + 5 − 7 + · · · = sin = 1. 2 2 2 3! 2 5! 2 7! Find the Maclaurin series of the function F (x) = May 23, 2011 x t e −1 dt. t 0 727

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